Unit 4 | unit 4 meromorphic functions and the logarithm

Unit - 4

Meromorphic Functions and the Logarithm

Q1) What is Residue Theorem?

A1)

We'd like to remind you of the following terms: Assume that f(z) is analytic at z0.

f(z) = an(z −z0)n + an+1(z −z0)n+1 + ...,

6 = 0 withan Then we say that f has an order n zero at z0. We call z0 a simple zero if n = 1. Assume that f has an isolated singularity at z0 and that the Laurent series exists.

Which culminates in 0 <|z −z0| < R and with bn=6 0. Then we say that f has an order n pole at z0. We call z0 a simple pole if n = 1.

Q2) Define Holomorphic and meromorphic. And state Behavior of functions near zeros and poles.

A2)

Holomorphic on A refers to a function that is analytic on a region A.

Meromorphic on A is a function that is analytic on A except for a collection of finite order poles.

Let

With (simple) poles at z = 2,3,4,5, this is meromorphic on C.

The essential assumption is that a function behaves like this near a zero of order n. (z −z0)n A function acts like this near a pole of order n. 1/(z −z0)n. This is made a little more precise by the following.

Behavior near a zero. If f has an order n zero at z0, then it is near z0.

f(z) ≈ an(z −z0)n,

For some constant an.

Proof. By definition, f has a Taylor series of the form around z0.

The claim follows because the second factor equals 1 at z0.

Behavior near a finite pole.

If f has a pole of order n at z0 then near z0,

For some constant bn.

Proof. This argument is virtually identical to the one before it. By definition, f has a Laurent series of the form around z0.

The claim follows because the second factor equals 1 at z0.

Picard’s theorem and essential singularities

Picard's theorem can be found near an essential singularity. This theorem will not be proved or used in 18.04. Nonetheless, we believe it is attractive enough to show you.

Picard’s theorem. If f(z) has an essential singularity at z0, then f(z) takes on all conceivable values infinitely many times in every neighbourhood of z0, with the possible exception of one value.

Q3) What is Quotients of functions?

A3)

The following is a statement concerning function quotients. If one or both functions had a pole instead of a zero, we might make similar statements.

Theorem. Suppose f has a zero of order m at z0 and g has a zero of order n at z0. Let

Then

● If n > m then h(z) has a pole of order n −m at z0. • If n < m then h(z) has a zero of order m −n at z0.

● If n = m then h(z) is analytic and nonzero at z0.

At z0, we can deduce that h(z) has a ‘pole' of order n -m. The ‘pole' is truly a zero if n -m is negative.

Proof. You should be able to provide proof of your claim. It's almost identical to the previous proofs:

Take the quotient by expressing f and g as Taylor series.

Let

We know that sin(z) has a first-order zero at z = 0 and z2 has a second-order zero. At z = 0, h(z) has a pole of order 1. Of course, using the Taylor series, we can clearly see this.

Q4) Define residues?

A4)

We'll look at how to calculate residues in this part. We've already seen enough to know that this will be beneficial. When we look at the residue theorem in the next part, we'll see this much more clearly.

In the last topic, we discussed residues. For the sake of thoroughness, we'll repeat the definition here.

Definition. Consider the function f(z) with an isolated singularity at z0, i.e. defined on the region 0 <|z −z0| < r and with Laurent series (on that region)

Is denoted Res(f,z0) = b1 or Res f = b1. z=z0

The residue of fat z0 is b1. This

What is the importance of the residue? If γis a small, simple closed curve that goes counter clockwise around b1 then

Γ small enough to be inside |z −z0| < r, surround z0 and contain no other singularity of f.

Integrating the Laurent series term by term demonstrates this. The phrase b1/z contains the only nonzero integral.

Example

At 0 has an isolated singularity Res(f,0) = 1/2, according to the Laurent series.

Q5) State properties of Residues at simple poles.

A5)

Simple poles occur frequently enough that we'll spend some time looking at how to compute their residues. There are several methods for locating a simple pole and calculating its residue. All of them have a reason that can be traced back to the Laurent series.

Assume that f(z) contains a singularity at z = z0. Then there are the properties listed below.

Property 1: If f(z Laurent )'s series is of the type

Then f has a simple pole at z0 and Res(f,z0) = b1.

Property 2 If g(z) = (z z0)f(z) at z0 is analytic, z0 is either a simple pole or a detachable singularity. Res(f,z0) = g in either case (z0). (In the event of a detachable singularity, the residue is zero.) Proof. For f near z0, directly from the Laurent series

Property 3: If f at z0 has a simple pole,

Lim (z −z0)f(z) = Res(f,z0) z→z0

This proves that the limit exists and that the residue equals the limit. If the limit is present, either the pole is simple or f is analytic at z0. The limit is equal to the residue in both circumstances.

Proof. For f near z0, directly from the Laurent series

Property 4. Res(fg,z0) = g(z0)Res(fg,z0) = g(z0)Res(fg,z0) = g(z0)Res(fg,z0) = g(z0)Res(fg,z0) = g(z0)Res(fg,z0) = g(z0) (f,z0).

If g(z0) 6 equals 0, then

Proof. Since z0 is a simple pole,

Since g is analytic,

g(z) = c0 + c1(z −z0) + ...,

When c0 equals g (z0). When these series are multiplied together, Res(fg,z0) = c0b1 = g(z0)Res (f,z0). QED

Because 1/g is analytic at z0, the statement concerning quotients f/g follows from the proof for products.

Property 5. 1/g(z) has a simple pole at z0 if g(z) has a simple zero at z0. And

Proof. The algebra for this is very similar to what we've done previously. For g, the Taylor expansion is

g(z) = a1(z −z0) + a2(z −z0)2 + ...,

Wherea1 = g0(z0). So

At z0, the second right-hand factor is analytic and equals 1. As a result, we know that 1/g's Laurent expansion is

Clearly the residue is 1/a1 = 1/g0(z0).

Q6) Explain Residues at finite poles with example.

A6)

We can make statements comparable to those for simple poles for higher-order poles, but the formulas and computations are more complicated. The following is the general principle: Poles of higher order. If at z0, f(z) has a pole of order k, then

g(z) = (z −z0)kf(z)

Is analytic at z0 and if

g(z) = a0 + a1(z −z0) + ...

Then

Q7) What is Residue at ∞?

A7)

The residue at ∞ is a clever mechanism that allows us to substitute the computation of several residues with the computation of a single residue in specific cases.

Assume that f is analytic in C with the exception of a small number of singularities. Let C be a big, positively oriented curve that contains all of the singularities.

All of the fare poles are contained inside C Definition. The fat infinite residue is defined as

We should start by explaining the concept. As you navigate a simple closed curve, everything to the left is the inside. Because curve C is counterclockwise, its interior contains all of f's poles. The residue theorem states that the residues of these poles define the integral over C.

The interior of the curve C, on the other hand, is everything outside of C. There are no fin poles in that area. The only choice is to have a residue at and define it as we did if we want the residue theorem to hold (which we do –it's that critical).

All of the poles of fare inside C are assumed in the definition of the residual at infinity. As a result, the residue theorem entails

Res(f,∞) = −X the residues of f.

We'll need a way to compute the residue directly to make this usable. The following theorem proves this.

Theorem. If f is analytic in C except for a finite number of singularities then

Res(f,∞) = −Res .

Proof. The proof is just a change of variables: w = 1/z.

Change of variable: w = 1/z

First note that z = 1/w and dz= −(1/w2)dw.

The map w = 1/z, on the other hand, transports the positively oriented z-circle of radius R to the negatively oriented w-circle of radius 1/R. (Observe how the circled points 1, 2, 3, 4 on C in the z-plane are transferred to points on C in the w-plane to see the orientation.) Thus,

Finally, all the poles inside the circle C are mapped to points outside the circle C by z = 1/w. So w = 0 is the only feasible pole of (1/w2)f(1/w) that is inside C. The residue theorem argues that because C is oriented clockwise,

Comparing this with the equation just above finishes the proof.

Example. Let

Earlier we computed

Calculate residues at z = 0 and z = 1 Calculate a single residue at infinity to recalculate this integral.

Solution:

We easily compute that

Res(f,∞) = −Res .

Since |z| = 2 contains all the singularities of fwe have

Q8) What are meromorphic functions?

A8)

A function on a domain Ω is called meromorphic, if there exists a sequence of points p1,p2,···with no limit point in Ω such that if we denote Ω∗= Ω \{p1,···}

● f : Ω∗→C is holomorphic.

● fhas poles at p1,p2···.

It is often useful to think of z = ∞ on the same footing as other points in the complex plane, and to define the extended complex plane

Cˆ = C∪∞.

We can then think of meromorphic functions f : Ω \{p1,···,pj,···}→C, as functions f : Ω →Cˆ, by defining

f(pj) = ∞

Pjs for all the poles Similarly, while considering meromorphic functions on C, it's also a good idea to think about the function's own extension to C. If z = 0 is a pole (resp. Detachable or essential singularity) for the function, we say z = is a pole (resp. Removable or essential singularity).

fˆ(z) = f(1/z).

If a meromorphic function on C does not have an essential singularity at z = ∞, we say it is meromorphic on the extended plane. Meromorphic functions on C can be classified, it turns out. Remember that a rational function on C is a form function.

Where both P(z) and Q(z) are polynomials.

Lemma 0.1. At infinity, a rational function possesses a pole or detachable singularity. If and only if degQ≥ degP, it possesses a detachable singularity.

Theorem 0.1. The only meromorphic functions on Cˆ are rational functions.

Proof. Let F :Cˆ →Cˆ be a meromorphic function.

Claim-1. F has only finitely many poles {p1,···,pn} in the complex plane C.

To see this, note that F(1/z) has either a pole or zero at z = 0. In either case there is a small neighborhood |z| <ε which has no other pole. Which is the same as saying that F has no finite pole in |z| >1/ε. But |z|≤ 1/ε is compact, and since all poles are isolated, this shows that there are only finitely many poles. Now, corresponding to each of the poles pk∈C there exists a polynomial Pk(see Remark 0.2 in Lecture-20) such that

Where Gk is holomorphic on pk's whole neighbourhood (including at the point pk). In the same way, we may write

Whereas before, G(z) is holomorphic in a neighborhood of z = 0. Claim number two is that the function

Is an entire and bounded function.

Assuming the claim, Liouville's theorem shows that H(z) is a constant, and so F(z) must be rational, proving the theorem. To demonstrate the assertion, first notice that H(z) is obviously holomorphic away from p1,•••,pn. At some z = pk, Pj(1/z −pj) is holomorphic for all j =6k. On the other hand, near pk,

That is, it is holomorphic. This demonstrates that H(z) is complete. As a result, we only need to show boundedness on |z| > R for some large R to demonstrate boundedness. To see this, remember that Pkare polynomials are polynomials.

Hence it is enough to show that F(z)−P∞(z) is bounded near infinity. But this follows immediately from noting that

Is holomorphic near z = 0 and hence is bounded on |z| <εfor some ε>0. In particular F(z) −P∞(z) is bounded on |z| >1/ε. This proves the claim, and hence completes the proof of the theorem.

The following theorem on partial fraction decomposition is a simple consequence of the proof, which we take for granted as an important tool in integration theory but never see the evidence of.

Corollary 0.1. For any rational function R(z) = P(z)/Q(z) has a partial fraction decomposition of the form

Where pk is a root of Q(z) of order mk, Pkis a polynomial of degree mk, and degP∞ = degP−degQ.

Q9) What are singularities?

A9)

A singularity is a point when an equation, surface, or other object blows up or degenerates. Singularities are also referred to as singular points.

Singularities play a crucial role in complex analysis, as they define the range of conceivable behaviours for analytic functions. Complex singularities are spots in a function's domain where the function fails to be analytic. Poles, essential singularities, logarithmic singularities, and detachable singularities are all types of isolated singularities. Natural boundaries or branch cuttings can produce nonisolated singularities.

Consider the ordinary differential equation of second order.

If and remain finite at , then is called an ordinary point. If either or diverges as , then is called a singular point. Singular points are further classified as follows:

1. If either or diverges as but and remain finite as , then is called a regular singular point (or nonessential singularity).

2. If diverges more quickly than , so approaches infinity as, or Q diverges more quickly than so that goes to infinity as , then is called an irregular singularity (or essential singularity).

A pole of order is a point of such that the Laurent series of has for and .

Essential singularities are poles of infinite order. A pole of order n is a singularity of for which the function is nonsingular and for which is singular for

A logarithmic singularity is a singularity of an analytic function whose main Z-dependent term is of order , , etc.

Removable singularities are singularities for which it is possible to assign a complex number in such a way that becomes analytic. For example, the function has a removable singularity at 0, since everywhere but 0, and can be set equal to 0 at. Removable singularities are not poles.

For example, the function has the following singularities: poles at , and a nonisolated singularity at 0.

Q10) What is argument principle?

A10)

The argument principle (or argument principle) is a result of the residue theorem. It links a curve's winding number to the number of zeros and poles within the curve. This is useful in applications (both mathematical and non-mathematical) when we need to know where the zeros and poles are located.

Principle of the argument

Setup.

A basic closed curve that is oriented counter clockwise.

Except for (potentially) some finite poles inside (not on) and some zeros inside (not on), f(z) is analytic on and inside.

Let p1, ...,pm be the poles of f inside γ.

Let z1, ...,zn be the zeros of f inside γ.

Write mult(zk) = the multiplicity of the zero at zk. Likewise write mult(pk) = the order of the pole at pk.

We begin with a theory that will lead to the principle of argumentation. Theorem 11.1 is the first of a series of theorems With the arrangement described above,

Proof. To prove this theorem we need to understand the poles and residues of f0(z)/f(z). With this in mind, suppose f(z) has a zero of order m at z0. The Taylor series for f(z) near z0 is f(z) = (z −z0)mg(z)

Whereg(z) is analytic and never 0 on a small neighborhood of z0. This implies

Since g(z) is never 0, g0(z)/g(z) is analytic near z0. This implies that z0 is a simple pole of f0(z)/f(z) and

Res= mult(z0).

Likewise, if z0 is a pole of order m then the Laurent series for f(z) near z0 is

f(z) = (z −z0)−mg(z)

Whereg(z) is analytic and never 0 on a small neighborhood of z0. Thus,

Again we have that z0 is a simple pole of f0(z)/f(z) and

Resmult(z0).

The following theorem comes directly from the Residue Theorem:

Sum of the residues

Definition. We write Zf,γfor the sum of multiplicities of the zerosof finside γ. Likewise for Pf,γ. So the Theorem 11.1 says,

(1)

Definition. Winding number. We have an intuition for what this means. We define it formally via Cauchy’s formula. If γis a closed curve then its winding number (or index) about z0 is defined as

Mapping curves: f ◦γ

One of the key notions in this topic is mapping one curve to another. That is, if z = γ(t) is a curve and w = f(z) is a function, then w = f ◦γ(t) = f(γ(t)) is another curve. We say f maps γto f ◦γ. We've done it before, but it's significant enough to us now that we'll pause and offer a few examples. This is a crucial notion in the reasoning principle, so make sure you understand it completely.

Example. Let γ(t) = eit with 0 ≤ t ≤ 2π (the unit circle). Let f(z) = z2. Describe the curve f ◦γ.

Solution: Clearly f ◦γ(t) = e2it traverses the unit circle twice as tgoes from 0 to 2π.

Example. Let γ(t) = it with −∞< t <∞ (the y-axis). Let f(z) = 1/(z + 1).

Describe the curve f ◦γ(t).

Solution: f(z) is a fractional linear transformation that transfers the line given by to the circle centred at 1/2 through the origin. By double-checking a few points:

We see that the circle is traversed in a clockwise manner as tgoes from −∞to ∞.

Im(z)Im(w)

The curve z = γ(t) = it is mapped to w = f ◦γ(t)) = 1/(it + 1).

Argument principle

This is also known as the principle of the argument.

Theorem. The premise of the argument With the same setup as before, for f

(2)

Proof. Theorem demonstrates that

As a result, we must show that the integral also equals the provided winding number. The change of variables w = f is all there is to it (z). With this change of variables the countour z = γ(t) becomes w = f ◦γ(t) and dw= f0(z)dzso

The preceding equation's final equivalence comes from the definition of a winding number.

Note that by assumption γdoes not go through any zerosof f, so w = f(γ(t)) is never zero and 1/w in the integral is not a problem.

Here is an easy corollary to the argument principle that will be useful to us later.

Corollary. Assuming that f ◦γdoes not go through −1, i.e. there are no zeros of 1+f(z) on γthen

. (3)

Proof. When we apply Equation 2's argument principle to the function 1 + f(z), we get

We can now compare each term in this equation to the terms in Equation 3:

(because (1 + f)0 = f0 )

Ind(1 + f ◦γ,0) = Ind(f ◦γ,−1) (1 + f winds around 0 ⇔f winds around -1)

Z1+f,γ= Z1+f,γ (same in both equations)

P1+f,γ= Pf,γ (poles of f = poles of 1 + f)

Example. Let f(z) = z2 + z Find the winding number of f ◦γ around 0 for each of the following curves.

γ1 = circle of radius 2.
γ2 = circle of radius 1 / 2.
γ3 = circle of radius 1.

Answers. f(z) has zeros at 0, −1. It has no poles.

So, f has no poles and two zeros inside γ1. The argument principle says Ind(f ◦γ1,0) = Zf,γ1 −Pf,γ= 2

Likewise f has no poles and one zero inside γ2, so Ind(f ◦γ2,0) = 1 −0 = 1

For γ3 a zero of fis on the curve, i.e. f(−1) = 0, so the argument principle doesn’t apply. The image of γ3 is shown in the figure below – it goes through 0.

The image of 3 different circles under f(z) = z2 + z.

Q11) Explain Rouche’s theorem.

A11)

Theorem. Theorem of Rouch. Consider the following scenarios:

Except for some finite poles, is a simple closed curve f,hare analytic functions on and within.

There are no poles of f and h on γ.

|h| <|f| everywhere on γ.

Then

Ind(f ◦γ,0) = Ind((f + h) ◦γ,0).

That is,

Zf,γ−Pf,γ= Zf+h,γ−Pf+h,γ (4)

Proof. We demonstrated a heuristic proof in class using a person walking a dog. f ◦γon a leash of length h ◦γ. Here is the analytic proof.

The argument principle requires the function to have no zeros or poles on γ. So we first show that this is true of f, f + h, (f + h)/f. The argument is goes as follows.

Zeros: The fact that 0 ≤|h| <|f| on γ implies f has no zeros on γ. It also implies f + h has no zeros on γ, since the value of h is never big enough to cancel that of f. Since f and f + h have no zeros, neither does (f + h)/f.

Poles: Assuming that f and h have no poles on, f +h also has no poles. (f + h)/f has no poles on because f has no zeros. The argument principle can now be used to f and f + h.

(5)

Ind((f + h) ◦γ,0) = Z f+h,γ−Pf+h,γ. (6)

Next, by assumption , so is inside the unit circle. This means that

Maps γ to the inside of the unit disk centered at 1. This implies that

Ind .

Let . The above says . (We showed above that g has no zeros or poles on γ.)

Now, it’s easy to compute that . So, using

Now equations 5 and 6 tell us Zf,γ−Pf,γ= Zf+h,γ−Pf+h,γ, i.e. we have proved Rouch´es theorem.

Corollary. Under the same hypotheses, If h and f are analytic (no poles) then

Zf,γ= Zf+h,γ.

Proof. Since the functions are analytic Pf,γand Pf+h,γare both 0. So Equation 4 shows Zf= Zf+h. QED.

We think of h as a small perturbation of f.

Example. Show all 5 zeros of z5 + 3z + 1 are inside the curve C2 :|z| = 2.

Solution: Let f(z) = z5 and h(z) = 3z + 1. Clearly all 5 roots of f (really one root with multiplicity 5) are inside C2. Also clearly, |h| <7 <32 = |f| on C2. The corollary to Rouch´es theorem says all 5 roots of f + h = z5 + 3z + 1 must also be inside the curve.

Example. Show z + 3 + 2ez has one root in the left half-plane.

Solution: Let f(z) = z + 3, h(z) = 2ez. Consider the contour from −iRto iRalong the y-axis and then the left semicircle of radius R back to −iR. That is, the contour C1 + CR shown below.

In order to apply the corollary to Rouch's theorem, we must first make sure that (for R large) |h| <|f| on C1 + CR. On C1, z = iy, so

|f(z)| = |3 + iy|≥ 3, |h(z)| = 2|eiy| = 2.

So |h| <|f| on C1.

On CR, z = x + iywith x <0 and |z| = R. So,

|f(z)| > R −3 for R large, |h(z)| = 2|ex+iy| = 2ex <2 (since x <0).

So |h| <|f| on CR.

The only zero of fis at z = −3, which lies inside the contour.

As a result of the Rouch es theorem's Corollary, f + h has the same number of roots as f inside the contour, which is 1. Allow R to reach infinity, then we can see that f + h has just one root in the half-plane.

Theorem. Fundamental theorem of algebra.

The following is how Rouch's theorem can be used to show the fundamental theorem of algebra.

Proof. Let

P(z) = zn+ an−1zn−1 + ... + a0

Be an nth order polynomial. Let f(z) = znand h = P −f. Choose an R such that

R >max(1,n|an−1|,...,n|a0|). Then on |z| = R we have

On |z| = R we have |f(z)| = Rn, so we have shown |h| <|f| on the curve. Thus, the corollary to Rouch´es theorem says f + h and f have the same number of zeros inside |z| = R. Since we know f has exactly n zeros inside the curve the same is true for the polynomial f + h. Now let R go to infinity, we’ve shown that f + h has exactly n zeros in the entire plane.

Note. The proof establishes a simple bound on the size of the zeros: they are all less than or equal to max in magnitude. (1,n|an−1|,...,n|a0|).

Q12) State the properties of the argument of a complex number.

A12)

Before we get started, let's go through the attributes of the argument of a non-zero complex number z, represented by argz (a multi-valued function), and the principal value of the argument, Argz, which is single-valued and typically defined as follows:

−π <Argz ≤ π (1)

The argument of a complex number is covered in detail in the class handout. We'll go through a few of the findings from that handout here. Argz can be thought of as a collection of the following elements:

Argz= Argz + 2πn, n = 0, ±1, ±2, ±3, ... , −π <Argz ≤ π . (2)

One can also express Argz in terms of argz as follows:

(3)

The biggest integer function is denoted by [ ]. That is, the greatest integer less than or equal to the real number x is defined as [x]. As a result, [x] is the only integer that fulfils the inequality.

x−1 <[x] ≤ x, for real x and integer [x]. (4)

For example, [1.5] = [1] = 1 and [−0.5] = −1. It is possible to verify that Argz, as defined in eq. (3), falls within the primary interval defined by eq (1). The properties of the multi-valued function argz are as follows:

Arg(z1z2) = argz1 + argz2 , (5)

(6)

(7)

Eqs. (5)–(7) should be regarded as set equalities, i.e. the elements of the sets represented by the left-hand and right-hand sides of the above identities are the same. The following outcomes, however, are not set equalities:

(8)

Which by vrtue of equations (5) and (6) yield

(9)

For example More generally,

(10)

We also take notice of various aspects of the argument's principal value.

(11)

(12)

Where the integers N± are determined as follows:

(13)

When we set z1 to 1 in eq. (12), we get

(14)

Note that for z real, both 1/z and z are also real so that in this case z = z and Arg(1/z) = Argz = Argz. In addition, in contrast to eq. (10), we have

Arg(zn) = nArgz + 2πNn, (15)

Where the integer Nnis given by:

(16)

And [ ] is the greatest integer bracket function introduced in eq. (4).

Q13) What are Properties of the real-valued logarithm, exponential and power functions?

A13)

Consider a positive real number's logarithm.

The quantity of properties:

This function fulfils a requirement.

x and y are positive real values, whereas a and p are arbitrary real numbers. Similarly, the power function defined over real numbers satisfies the following conditions:

x and y are positive real values, whereas a and b are random real numbers. The generalised exponential function defined over real numbers is closely connected to the power function. This function fits the following criteria:

For positive real numbers a and b and arbitrary real numbers x and y.

We would like to know which of these relations are satisfied when these functions are extended to the complex plane. It is dangerous to assume that all of the above relations are valid in the complex plane without modification, as this assumption can lead to seemingly paradoxical conclusions. Here are three examples:

1. Since 1/(−1) = (−1)/1 = −1 ,

. (37)

Hence, 1/i= ior i2 = 1. But i2 = −1, so we have proven that 1 = −1.

Since 1 = (−1)(−1) ,
. (38)

To prove that ln(−z) = ln(z) for all z =6 0, we proceed as follows:

Ln(z2) = ln[(−z)2],

Ln(z) + ln(z) = ln(−z) + ln(−z),

2ln(z) = 2ln(−z), ln(z) = ln(−z).

All of these "proofs" are, of course, false. The first two proofs contain a fallacy due to eqs. (28) and (29), which are true for real-valued functions but not for complex-valued functions in general. The third proof's mistake is more subtle, and it will be addressed later in these comments. A thorough understanding of the complex logarithm, power, and exponential functions will explain how to adjust eqs. (17)–(36) correctly and avoid common mistakes that can lead to incorrect results.

Q14) Define the complex exponential function and complex logarithm.

A14)

We'll start with the complex exponential function, which has the following power series:

Where z is a complex number of any type. Using this definition of a power series, one may verify that:

, for all complex and (39)

If z = x + iy, where x and y are real numbers, it follows that

ez= ex+iy= ex eiy= ex(cosy + isiny).

The complex exponential function can be used to quickly verify that eqs. (30)–(33) are met. Furthermore, when the outer exponent is an integer, eq. (34) holds:

(ez)n = enz, n = 0, ±1, ±2, ... . (40)

The resulting equation is a multivalued power function if the outer exponent is a non-integer. In section 8, we'll go over this case in further depth.

Before we continue, there is one important property of the complex exponential to note:

e2πin = 1, n = 0, ±1, ±2, ±3, ... . (41)

Definition of the complex logarithm

The complex logarithm must be defined by solving the complex equation:

z = ew, (42)

Forw, where z is any complex number that is not zero. If we write w = u + iv, we may write eq. (42) as

Eueiv= |z|eiargz . (43)

Eq. (43) implies that:

|z| = eu, v = argz .

The equation |z| = euis a real equation, so we can write u = ln|z|, where ln|z| is the ordinary logarithm evaluated with positive real number arguments. Thus,

w = u + iv = ln|z| + iargz= ln|z| + i(Argz + 2πn), n = 0, ±1, ±2, ±3, ... (44)

The complex logarithm is denoted by the symbol w, which is written as w = lnz. Since we have used the symbol ln for the real logarithm in eq. (44) this is an uncomfortable notation. The real logarithm in eq. (44) will be denoted by the symbol Ln to solve this notational conundrum. That is, the ordinary real logarithm of |z| will be denoted as Ln|z|. We may rephrase eq. (44) using this notational convention:

Lnz= Ln|z| + iargz= Ln|z| + i(Argz + 2πn), n = 0, ±1, ±2, ±3, ... (45)

For any complex number z that is not zero

Lnzi is clearly a multi-valued function (as its value depends on the integer n). It is useful to define a single-valued complex function, Ln z, called the principal value of lnzas follows:

Lnz= Ln|z| + iArgz , −π <Argz ≤ π , (46)

Which extends the definition of Ln z to the entire complex plane (excluding the origin, z = 0, where the logarithmic function is singular). In particular, eq. (46) implies that Ln(−1) = iπ. Note that for real positive z, we have Argz = 0, so that eq. (46) simply reduces to the usual real logarithmic function in this limit. The relation between lnzand its principal value is simple:

Lnz= Ln z + 2πin, n = 0, ±1, ±2, ±3, ... .

Q15) State properties of the complex logarithm.

A15)

Now we'll look at whether of the properties in equations (17)–(22) apply to the complex logarithm. We should look at the properties of both the multi-valued function lnz and the single-valued function Ln z now that we've defined them. Let's start with the lnz multi-valued function. First, we'll look at eq (17). Using eq. (45), we can deduce:

Elnz= eLn|z|eiArgze2πin = |z|eiArgz = z . (47)

Thus, eq. (17) is satisfied. Next, we examine eq. (18) for z = x + iy:

Ln(ez) = Ln|ez| + i(argez) = Ln(ex) + i(y + 2πk) = x + iy+ 2πik= z + 2πik ,

Wherek is an arbitrary integer. In deriving this result, we used the fact that ez= exeiy, which implies that arg(ez) = y + 2πk.‡ Thus,

Ln(ez) = z + 2πik=6 z, unless k = 0. (48)

This is not surprising, since ln(ez) is a multi-valued function, which cannot be equal to the single-valued function z. Indeed eq. (18) is false for the multi-valued complex logarithm.

As a check, let us compute ln(elnz) in two different ways. First, using eq. (47) , it follows that ln(elnz) = lnz. Second, using eq. (48), ln(elnz) = lnz+ 2πik. This seems to imply that lnz= lnz+ 2πik. In fact, the latter is completely valid as a set equality in light of eq. (45).

Now we'll look at the properties shown in eqs. (19)–(20). (22). Equations (19)–(21) are satisfied as set equalities using the definition of multi-valued complex logarithms and the properties of argz given in eqs. (5)–(7):

However, one must be careful in employing these results. One should not make the mistake of writing, for example, lnz+ lnz=? 2lnz or lnz−lnz=? 0. For the same reason that eqs. (8) and (9) are not identities under set equality, both of these claims are untrue. When p is an integer n, the multi-valued complex logarithm does not meet eq. (22):

Which is inferred from eq (10). If p is not an integer, zp is a complex multivalued function, then the validity of eq. (22) requires more investigation. In section 6, we'll show [see eq. (62)] that the complex logarithm can only satisfy eq. (22) if p = 1/n, where n is an integer. In this instance,

The properties of the single-valued function Ln z are next investigated. We look at the six qualities given by eqs. (17)– (19) once again (22). For starters, eq. (17) is easily met since

However, eq. (18) is frequently incorrect. Given example, for z = x + iy

After using eq. (3), where [ ] is the greatest integer bracket function defined in eq. (4). Thus, eq. (18) is satisfied only when −π < y ≤ π. For values of y outside the principal interval, eq. (18) contains an additive correction term as shown in eq. (55).

As a check, let us compute Ln(eLnz) in two different ways. First, using eq. (54) , it follows that Ln(eLnz) = Ln z. Second, using eq. (55) ,

Where we have used Im Ln z = Argz [see eq. (46)]. In the last step, we noted that

Due to eq. (1), which implies that the integer part of Argz is zero. Thus, the two computations agree.

Now we'll look at the properties shown in eqs. (19)–(20). (22). Because the complex logarithm's primary value must be within the interval, Ln z may not meet any of these requirements. −π <Im Ln z ≤ π. Using the results of eqs. (11)–(16), it follows that

Ln (z1z2) = Ln z1 + Ln z2 + 2πiN+ , (56)

Ln (z1/z2) = Ln z1 −Ln z2 + 2πiN−, (57)

Ln(zn) = nLnz + 2πiNn (integer n), (58)

Where the integers N± = −1, 0 or +1 and Nnare determined by eqs. (13) and

(16), respectively, and

(59)

Note that eq. (19) is satisfied if Re z1 >0 and Re z2 >0 (in which case N+ = 0). In other cases, N+ =6 0 and eq. (19) fails. Similar considerations also apply to eqs. (20)–(22). For example, eq. (21) is satisfied by Ln z unless Argz = π (equivalently for negative real values of z), as indicated by eq. (59). In particular, one may use eq. (58) to verify that:

Ln[(−1)−1] = −Ln(−1) + 2πi= −πi+ 2πi= πi= Ln(−1),

As expected, since (−1)−1 = −1.

We can't examine whether eq. (22) holds if p is a non-integer because zp is a multi-valued function in this situation. As a result, we'll now focus on the complicated power functions (and the related generalised exponential functions).

Q16) Define the generalized power and exponential functions.

A16)

The equation for the generalised complex power function is as follows:

To justify this definition, consider that if c = k is an integer, then z = |z|eiargz is an integer.

Zk= |z|kekiargz= ekLn|z|ekiargz= ek(Ln|z|+iargz) = eklnz .

In this case, w = zkis a single-valued function, since

Zk= eklnz= ek(Ln z +2πin) = ekLn z .

If c = 1/k (where k is an integer), then we have:

z1/k= |z|1/keiarg(z)/k = eLn|z|/keiarg(z)/k = e(Ln|z|+iargz)/k = eln(z)/k ,

Where|z|1/k refers to the real kth root of |z| that is positive. We can easily prove that eq. (60) holds for any rational real number c by combining the two conclusions just obtained. Because any irrational real number can be approximated (to any desirable degree of precision) by a rational number, eq. (60) must hold for any real number c. These considerations motivate the use of eq. (60) to define the generalised complex power function for any complex power c.

Note that due to the multi-valued nature of lnz, it follows that w = zc= eclnzis also multi-valued for any non-integer value of c, with a branch point at z = 0:

w = zc= eclnz= ecLnze2πinc, n = 0, ±1, ±2, ±3,···. (61)

If c is a rational number, then it can always be expressed in the form c = m/k, where m is an integer and k is a positive integer such that m and k possess no common divisor. One can then assume that n = 0,1,2,...,k−1 in eq. (61), since other values of n will not produce any new values of zm/k. It follows that the multi-valued function w = zm/k has precisely k distinct branches. If c is irrational or complex, then the number of branches is infinite (with one branch for each possible choice of integer n).

Having defined the multi-valued complex power function, we are now able to compute ln(zc). In light of eq. (48) ,

(62)

Where k is an arbitrary integer. Thus, ln(zc) = clnz in the sense of set equality ( in which case the sets corresponding to lnz and lnz+2πik/c coincide) if and only if k/c is an integer for all values of k. Taking c = 1/n, where n is an integer, is the only way to satisfy this criterion. As a result, eq. (53) has been verified.

We can define a single-valued power function by selecting the principal value of lnzin eq. (60). Consequently, the principal value of zcis defined by

For lack of a better notation, I'll capitalise the variable Z as shown above to represent the principal value. The principal value definition of zccan lead to some unexpected results. For example, consider the principal value of the cube root function w = Z1/3 = eLn(z)/3. Then, for z = −1, the principal value of

This may have surprised you, if you were expecting that. To obtain the latter result would require a different choice of the principal interval in the definition of the principal value of z1/3.

In the case where both the complex logarithm and the complex power function are specified by their principal values, we can now check eq. (22). To put it another way, we compute:

Ln(Zc) = Ln(ecLnz) = cLnz+ 2πiNc, (64)

Following the application of eq. (55), where Ncis is an integer defined by

, (65)

The biggest integer bracket function defined in eq. Is [ ]. (4). It is possible to assess Nc by noticing that:

Im (cLnz) = Im{c(Ln|z| + iArgz)} = Argz Re c + Ln|z|Imc.

It's worth noting that if c = n and n is an integer, eq. (64) simplifies to eq. (58) as expected. We conclude that both the multi-valued complex logarithm and its primary value are erroneous in eq. (22).

The generalised exponential function is a function that is comparable to the complicated power function in several ways. The generalised exponential function for c =6 0 could be defined as follows:

(66)

The multivalued nature of this function, however, differs from the multivalued power function in some ways. The generalised exponential function, in contrast to the latter, has no branch point (or any other sort of singularity) in the finite complex z-plane. For each value of n, eq. (66) can be thought of as creating a set of independent single-valued functions. The n = 0 case is usually the most practical, in which case we simply define:

This matches our concept of the exponential function in section 3 (c = e). We'll use eq. (67) as the single-valued generalised exponential function definition from now on. §

Some of the results for the complex power function's principal value can be applied to the generalised exponential function right away. For example, a nearly identical computation as in eqs. (64) and (65) yields the following results:

(68)

Where is an integer determined by:

Q17) State Properties of the generalized power function.

A17)

Let's take a look at the properties listed in eqs (29). The complex power function is defined by Eq. (23) It's all too easy to write:

However, consider the case of non-integer a andb where a + b is an integer. In this case, eq. (70) cannot be correct since it would equate a multi-valued function zazb with a single-valued function za+b. In fact, the questionable step in eq. (70) is false:

In the case of a = b = 1 [cf. Eq. (8)], we already know that eq. (71) is untrue. A more thorough calculation yields:

Where k and n are arbitrary integers. Hence, za+bis a subset of zazb. Whether the set of values for zazband za+bdoes or does not coincide depends on a andb.

However, in general, eq. (24) does not hold.

Similarly,

(73)

In practice, many textbooks treat the generalized exponential function as a single-valued function, cz= ezLn c, only when c is a positive real number. For any other value of c, the multi-valued function cz= ezlncis preferred. We shall not pursue this approach in these notes.

Where k and n are arbitrary integers. Hence, za−b is a subset of za/zb. Whether the set of values za/zband za−b does or does not coincide depends on a and b. However, in general, eq. (25) does not hold. Setting a = b in eq. (73) yields the expected result:

z0 = 1, z =6 0

For any complex number z that is not zero The set equality is obtained by setting a = 0 in eq. (73):

(74)

i.e., the set of values for z−b and 1/zbcoincide. Thus, eq. (26) is satisfied. Note, however, that

Zaz−a = ealnze−alnz= ea(lnz−lnz) = ealn1 = e2πika ,

Wherek is an arbitrary integer. Hence, if a is a non-integer, then zaz−a =6 1 for k =6 0. This is not in conflict with the set equality given in eq. (74) since there always exists at least one value of k (namely k = 0) for which zaz−a = 1. To show that eq. (27) can fail, we use eq. (48) to conclude that

(75)

Where k is a random number. When we use the principal value of the generalised power function, which is described in eq. (63), we get eqs. (61) and (75).

k and n are both arbitrary integers. Thus, zabis a subset of (za)b. Note that the elements of zaband (za)b coincide if a = ±1 and/or b is an integer. In contrast, if z = a = b = i, then zab= −i, whereas eq. (75) yields,

The multi-valued power function, on the other hand, satisfies eqs. (28) and (29) because

Other special cases exist in which the elements of zab and (za)b coincide. For example, if a = 3/2 and b = 1/2, then one can check that allowing for all possible integer values of n and k in eqs. (76) and (77) yields (za)b= zab= {Zab,−Zab,iZab,−iZab}.

We now repeat the above analysis for the principal value of the power function, Zc= ecLnz. In this case, the results are somewhat reversed from the case of the multi-valued power function. In particular, eqs. (24)–(26) are satisfied, whereas eqs. (27)–(29) may be violated. For example, for the single-valued power function,

(79)

(80)

As expected, setting a = b in eq. (80) results in Z0 = 1 (for z 6=0). Since eq. (55) implies that, eq. (27) may be broken.

(Zc)b= (ecLnz)b = ebLn(ecLnz) = eb(cLnz+2πiNc) = ebcLnz e2πibNc= Zcb e2πibNc

Where Ncis an integer determined by eq. (65). As an example, if z = b = c = i, eq. (65) gives Nc= 0, which yields the principal value of (ii)i= ii·i= i−1 = −i. However, in general Nc=6 0 is possible in which case (Zc)b6=Zcb[i.e., eq. (27) does not hold] unless bNcis an integer. For example, if z is real and negative and c = −1, then Nc= 1 and (Z−1)b= Z−be2πib[which provides a resolution for the paradox of eq. (37)]. That is, if z is real and negative then (Z−1)b6=Z−b unless b is an integer.

Eqs. (28) and (29) may also be violated since eqs. (56) and (57) imply that

Where the integers N± are determined

Properties of the generalized exponential function

The single-valued generalised exponential function w = cz(c =6 0) is defined by eq (67). One can rapidly examine whether hold in the complex plane using this formulation and the features of the complex exponential function ez. Are virtually identical to in terms of proof:

However, eq. (34) does not generally hold. Using eq. (68) ,

(cz1)z2= ez2 Ln(cz1) = ez2(z1 Ln c+2πiNc′) = ez2z1 Ln c e2πiz2Nc′= cz1z2 e2πiz2Nc′, (87) where is determined by eq. (69) [with z replaced by z1]. The case of c = e is noteworthy. Eq. (87) reduces to:

(88)

If z2 = n where n is any integer, then e2πinNe′= 1 (since is an integer by definition of the bracket notation). Thus, we recover eq. (40).

Then, = 1 and hence (e−πi)z = eiπz .

Let us test eq. (88) by substituting z1 = −iπ

This result may seem strange, but it is a consequence of our definition of the generalized exponential function, cz= ezLn c, which employs the principal value of the logarithm. Indeed

Since Ln(−1) = iπ. We conclude that eq. (34) can be violated, even for the ordinary exponential function.

Ultimately, the real difficulty with (cz1)z2is that it is simultaneously a generalized exponential function and a generalized power function. Thus, if z2 is a non-integer, it may be more convenient to treat (cz1)z2as a multi-valued function. That is, in this latter convention, we treat the generalized exponential function cz1 = ez1 Ln c as a single-valued function (using the principal value definition of the logarithm in the exponent), whereas we treat the generalized power function (cz1)z2= ez2 ln(cz1) as a possible multi-valued function:

(cz1)z2= ez2 ln(cz1) = ez2 ln(ez1 Ln c) = ez2(z1 Ln c+2πik) = ez2z1 Ln c e2πiz2k = cz1z2 e2πikz2 ,

Where k is an arbitrary integer [see eq. (48)]. In particular, for z2 a non-integer, (cz1)z2is a multi-valued function, with branches corresponding to different choices of k. For example, for c = z1 = z2 = i, we recover eq. (78). One might be tempted to call the k = 0 branch the principal value of (cz1)z2, in which case eq. (34) would be valid. Clearly, we must define our conventions carefully if we wish to manipulate expressions involving exponentials of exponentials.

Finally, the equations (35) and (36) may be broken. The procedure is roughly identical to that described in equations (82) and (83):

Where N ± is calculated by eq. (13) [where z1 and z2 are replaced by a and b, respectively]. N = 0, and eqs. (35) and (36) are met if Re a >0 and Re b >0.

Q18) Evaluate

A18)

Singularities have been isolated at z = 0,±i and a zero at z = −1. We will show that z = 0 is a pole of order 3, z = ±iare poles of order 1 and z = −1 is a zero of order 1. The style of argument is the same in each case.

At z = 0: