Unit – 5
Unit – 5
Algebra of Complex numbers and Roots of polynomial Equations
Q 1: Simplify 16 i+10i (3-i)
Solution:
Given,
16i +10i (3-i)
=16i+10i(3i) +10i(-i)
=16i+30i-10i2
=46i-10(-1)
=46i+10
Here real part is 10 and imaginary part is 46
Q 2: Express the following into a+ib form
Solution:
Given., 
z = 
= 
=
+
i
Modulus, 
= 
=
= 
Conjugate
= (
- 
Q 3: Find the polar form of 7-5j
Solution:
We need to find r and 
r=
=
= 
=
8.6
To find θ, we first find the acute angle α ,
= tan-1(
)
= tan-1(
)
= 35.540
7-5j is in the fourth quadrant so we we have,
= 3600 -35.540 = 324.460
So, expressing 7-5j in the polar form as.,
7-5j = 8.6(cos324.50 +j sin324.50)
Q 4: Represent 1+j
in polar form
= arctan 
=600
Solution: R= 
= 
=2
So,
1+j
= 2
600 = 2(cos600 +jsin 600)
Q 5: Write ( 
) 7 in the form of a+ib
First determine the radius ie., r
Solution: Let r= 
r = 
+ i2
r= 
r=2
Now cos
=-
and sin 
=
, 
must be in the first quadrant and 
=300
Therefore,
( 
) 7 = 
7
( 
) 7 = 27 [cos 7.300 +i sin 7.300]
= 128[cos 2100+isin2100]
= 28
+i 
( 
) 7 = 64
-64 i
Q 6: Write (
+i 
)4 in the form of a+ib
Solution: First, we determine the radius,
r = 
r= 
r=
r=2
Now,
(
+i 
)4 = 
4
= [24 (cos 4.3150 + isin 4.3150)
=16(cos 21600+i sin 21600)
= 16( cos 1800+i.sin 1800)
= 16(-1+0i)
= -16
Q 7: Prove that function 
is analytic function.
Solution: Real and Imaginary parts of 
are
If, 
On differentiating u,v we get
Again differentiating
Hence e-R-Equation satisfies.
Q 8: 2y=4x−16 and y=2x−8 perpendicular.
Solution: Let 2y =4x-16 
(1)
Y=2x-8 
(2)
Let s write equation (1) in the form of y= mx+c and 
2)
y= 2x-8 and y = 2x-8
Here we see that there is only one line,
But if there are two lines, they would have been parallel as the gradients are same.
y= 2x+5 and y= 2x-8 are parallel
y= --
x-8 and y=2x-8 are perpendicular
--
are perpendicular.
Q 9: Lisa will make up a punch that is 25% fruit juice by adding pure fruit-juice to a
2-liter mixture that is 10% pure fruit-juice. How many litres of pure fruit juice does she need to add?
Solution:
Let’s call the amount to be found z
Then you will end up with x+2L of 25% juice
This will contain 0.25(x+2) =0.25x+0.5pure juice
The original 2L already contained 0.10.2=0.2 juice
So we added 0.25x=0.3 juice, but this is also x(as x=100% juice).
0.25x+0.3=x
0.75x=0.3 
x=0.4 liter
Q10: Finding the real root of the following polynomial equation using cardano’s method
X3+2x2+3x+4=0
Solution: Given that., a=1, b=2, c=3, d=4
Compute Q and R:
Q=
= 
R= 
=
Compute S and T :
S=
= 
T= 
= 
Compute the roots:
X1 = S+T- 
= 
= -1.651
X2 = 
-
+i
(s-t)
=
+ i.
X3 = 
-
+i
(s-t)
- i.
-0.175 -1.547 i.
Given quadratic equation: 36x4 -72x3-391x2-123x+270
Applying Ferrari’s method:
We have,
A=36, B=-72, C=-391, D=-123, E=270
- 
=- 
- 
+
= -
-
-
+
= 
P= - 
–
=
=15.62
16
Q= 
- 
-
= - 
=23.59
24
Finally,
R= -
= -
By solving the above equation, we get., 12
.
Q 11: Consider the quadratic equation y4+ay3+by2+cy+d=0
Solution: By substituting y= x-
the above equation will be reduced to the following
=X4+ px2+qz+r=0,
(1)
The above equation doesn’t contain x3., if one introduces an auxiliary parameter
,then the left hand side of equation (1) can be written as follows:
=X4+ px2+qz+r = (x2 + 
+
)2 – [2ax2 –qx+(
2 +p
+
-r)]
(2)
= q2 -4.2
(
2 +p
+
–r) =0
Then the required equation for 
=
0 the polynomial in square brackets in equation (2) has one double root
X0= 
Unit – 5
Algebra of Complex numbers and Roots of polynomial Equations
Q 1: Simplify 16 i+10i (3-i)
Solution:
Given,
16i +10i (3-i)
=16i+10i(3i) +10i(-i)
=16i+30i-10i2
=46i-10(-1)
=46i+10
Here real part is 10 and imaginary part is 46
Q 2: Express the following into a+ib form
Solution:
Given., 
z = 
= 
=
+
i
Modulus, 
= 
=
= 
Conjugate
= (
- 
Q 3: Find the polar form of 7-5j
Solution:
We need to find r and 
r=
=
= 
=
8.6
To find θ, we first find the acute angle α ,
= tan-1(
)
= tan-1(
)
= 35.540
7-5j is in the fourth quadrant so we we have,
= 3600 -35.540 = 324.460
So, expressing 7-5j in the polar form as.,
7-5j = 8.6(cos324.50 +j sin324.50)
Q 4: Represent 1+j
in polar form
= arctan 
=600
Solution: R= 
= 
=2
So,
1+j
= 2
600 = 2(cos600 +jsin 600)
Q 5: Write ( 
) 7 in the form of a+ib
First determine the radius ie., r
Solution: Let r= 
r = 
+ i2
r= 
r=2
Now cos
=-
and sin 
=
, 
must be in the first quadrant and 
=300
Therefore,
( 
) 7 = 
7
( 
) 7 = 27 [cos 7.300 +i sin 7.300]
= 128[cos 2100+isin2100]
= 28
+i 
( 
) 7 = 64
-64 i
Q 6: Write (
+i 
)4 in the form of a+ib
Solution: First, we determine the radius,
r = 
r= 
r=
r=2
Now,
(
+i 
)4 = 
4
= [24 (cos 4.3150 + isin 4.3150)
=16(cos 21600+i sin 21600)
= 16( cos 1800+i.sin 1800)
= 16(-1+0i)
= -16
Q 7: Prove that function 
is analytic function.
Solution: Real and Imaginary parts of 
are
If, 
On differentiating u,v we get
Again differentiating
Hence e-R-Equation satisfies.
Q 8: 2y=4x−16 and y=2x−8 perpendicular.
Solution: Let 2y =4x-16 
(1)
Y=2x-8 
(2)
Let s write equation (1) in the form of y= mx+c and 
2)
y= 2x-8 and y = 2x-8
Here we see that there is only one line,
But if there are two lines, they would have been parallel as the gradients are same.
y= 2x+5 and y= 2x-8 are parallel
y= --
x-8 and y=2x-8 are perpendicular
--
are perpendicular.
Q 9: Lisa will make up a punch that is 25% fruit juice by adding pure fruit-juice to a
2-liter mixture that is 10% pure fruit-juice. How many litres of pure fruit juice does she need to add?
Solution:
Let’s call the amount to be found z
Then you will end up with x+2L of 25% juice
This will contain 0.25(x+2) =0.25x+0.5pure juice
The original 2L already contained 0.10.2=0.2 juice
So we added 0.25x=0.3 juice, but this is also x(as x=100% juice).
0.25x+0.3=x
0.75x=0.3 
x=0.4 liter
Q10: Finding the real root of the following polynomial equation using cardano’s method
X3+2x2+3x+4=0
Solution: Given that., a=1, b=2, c=3, d=4
Compute Q and R:
Q=
= 
R= 
=
Compute S and T :
S=
= 
T= 
= 
Compute the roots:
X1 = S+T- 
= 
= -1.651
X2 = 
-
+i
(s-t)
=
+ i.
X3 = 
-
+i
(s-t)
- i.
-0.175 -1.547 i.
Given quadratic equation: 36x4 -72x3-391x2-123x+270
Applying Ferrari’s method:
We have,
A=36, B=-72, C=-391, D=-123, E=270
- 
=- 
- 
+
= -
-
-
+
= 
P= - 
–
=
=15.62
16
Q= 
- 
-
= - 
=23.59
24
Finally,
R= -
= -
By solving the above equation, we get., 12
.
Q 11: Consider the quadratic equation y4+ay3+by2+cy+d=0
Solution: By substituting y= x-
the above equation will be reduced to the following
=X4+ px2+qz+r=0,
(1)
The above equation doesn’t contain x3., if one introduces an auxiliary parameter
,then the left hand side of equation (1) can be written as follows:
=X4+ px2+qz+r = (x2 + 
+
)2 – [2ax2 –qx+(
2 +p
+
-r)]
(2)
= q2 -4.2
(
2 +p
+
–r) =0
Then the required equation for 
=
0 the polynomial in square brackets in equation (2) has one double root
X0= 
