EQUALITY OF TWO COMPLEX NUMBERS

Problem 1 :

Find real numbers a and b such that :

a)  a + ib = 4 

b) (1 – 2i) (a + bi) = -5 – 10i

c)  (a + 2i) (1 + bi) = 17 – 19i

Solution :

a)  a + ib = 4

By equating the real and imaginary parts.

a = 4 and b = 0

b)  (1 – 2i) (a + bi) = -5 – 10i

By multiplying the binomials, we get

a + bi - 2ia - 2bi2 = -5 - 10i

Here the value of i2 is -1.

a + bi - 2ia + 2b = -5 - 10i

(a + 2b) + i(b - 2a) = -5 - 10i

By equating the real and imaginary parts.

a + 2b = -5 ---(1)

b - 2a = -10

-2a + b = -10  ----(2)

(1) ⋅ 2 + (2)

2a + 4b - 2a + b = -10 - 10

5b = -20

b = -4

Applying the value of b in (1), we get

a - 8 = -5

a = 3

c.

(a + 2i) (1 + bi) = 17 – 19i

By multiplying two complex numbers on the left side, we get

a + abi + 2i + 2bi2 = 17 - 19i

Here the value of i2 is -1.

a + abi + 2i - 2b = 17 - 19i

(a - 2b) + i(ab + 2) = 17 - 19i

By equating the real and imaginary parts.

a - 2b = 17

a = 17 + 2 b --- (1)

ab + 2 = -19

ab = -21  --- (2)

Applying the value of a in (2), we get

(17 + 2b)b = -21

17b + 2b2 = -21

2b2 + 17b + 21 = 0

(b + 7)(2b + 3) = 0

b = -7, b =  -3/2

When b = -7

a = -21/(-7)

a = 3

When b = -3/2

a = -21/(-3/2)

a = 14

Problem 2 :

Find a complex number z such that 2z -1 = iz – i. Write your answer in the form z = a + bi where a, b ϵ ℝ.

Solution :

2z -1 = iz – i

2z - iz = -i + 1

z(2 - i) = -i + 1

Dividing 2 - i on each sides.

z = -i + 12 - i= -i + 12 - i 2 + i2 + i= -2i - i2 + 2 + i4 + 2i - 2i - i2= -2i + 1 + 2 + i4 + 1= 3 - i5= 35 - i5

Problem 3 :

if z = 4 + i and w = 3 – 2i find 2w* - iz.

Solution :

Given, z = 4 + i and w = 3 - 2i

To find 2w* - iz :

w* = 3 + 2i

= 2(3  +2i) - i(4 + i)

= 6 + 4i - 4i - i2

= 6 + 1

= 7

Problem 4 :

Find rationals a and b such that

(2 – 3i)/(2a + bi) = 3 + 2i.

Solution :

Given (2 – 3i)/(2a + bi) = 3 + 2i

Multiplying 2a + bi on each sides.

2 – 3i = (3 + 2i) (2a + bi)

2 - 3i = 6a + 3bi + 4ai + 2bi2

2 - 3i = 6a + 3bi + 4ai - 2b

2 - 3i = 6a - 2b + i(3b  + 4a)

Equating real and imaginary parts.

6a - 2b = 2

3a - b = 1 --- (1)

4a + 3b = -3 -----(2)

(1) ⋅ 3 + (2)

9a - 3b + 4a + 3b = 3 - 3

13a = 0

a = 0

Applying the value of a in (1), we get

3(0) - b = 1

-b = 1

b = -1

Problem 5 :

a + ai is a root of x2 – 6x + b = 0 where a, b ϵ ℝ. Explain why b has two possible values. Find a in each case.

Solution :

Since one of the root is in the form of complex number, other root will be its conjugate.

So, the roots are a + ai and a - ai

Sum of the roots :

= (a + ai) + (a - ai)

=   a + ai + a - ai

= 2a

Products of the roots :

= (a + ai) (a - ai)

= a2 + a2

= 2a2

x2 - (sum of the roots)x + products of the roots = 0

x2 - 2ax + 2a2 = 0

Given x2 - 6x + b = 0

x2 - 2ax + 2a2 = x2 - 6x + b

Equating coefficients of x

-2a = -6

a = -6/-2

a = 3

Equating constants terms

2a2 = b

2(3)2 = b

18 = b

Problem 6 :

Find real numbers x and y such that

(3x + 2yi) (1 - i) = (3y + 1)i – x.

Solution :

(3x + 2yi) (1 - i) = (3y + 1)i – x

3x - 3xi + 2yi - 2yi2 = 3yi + i - x

3x - 3xi + 2yi + 2y = 3yi + i - x

(3x + 2y) + i (2y - 3x) = i(3y + 1) - x

By equating the real and imaginary parts.

3x + 2y = -x

2y = -x - 3x

-4x - 2y = 0

2x + y = 0 ---(1)

2y - 3x = 3y + 1

-3x - y = 1--- (2)

(1) + (2)

2x + y - 3x - y = 0 + 1

-x = 1

x = -1

By applying the value of x in (1), we get

-2 + y = 0

y = 2

Problem 7 :

Find real x and y such that :

a) x + iy = 0

b) (3 – 2i) (x + i) = 17 + yi 

c) (x + iy)2 = x – iy

Solution :

a) 

x + iy = 0

Equating real and imaginary parts.

x = 0 and y = 0

b) 

(3 – 2i) (x + i) = 17 + yi

3x + 3i - 2ix + 2 = 17 + yi

(3x + 2) + i(3 - 2x) = 17 + yi

Equating real and imaginary parts.

3x + 2 = 17 --- (1)

x = 5

3 - 2x = y --- (2)

x = 5 substitute the equation (2).

3 - 2(5) = y

3 - 10 = y

-7 = y

So, the real numbers x and y is 5 and -7.

c)

(x + iy)2 = x – iy

x2 + i2y2 + 2(x)(iy) = x - iy

x2 - y2 + 2xiy = x - iy

Equating real and imaginary parts.

x2 - y2 = x --- (1)

2xy = -y --- (2)

2x = (-y) (1/y)

2x = -1

x = -1/2

x = -1/2 substitute the equation (1).

-122- y2 = x14 - y2 = -12-y2 = -12 - 14-y2 = -2 - 14-y2 = -34y = 34y = 32

Problem 8 :

Find z if √z = 2/(3 – 2i) + 2 + 5i

Solution :

z = 23- 2i + 2 + 5i= 23- 2i + (2 + 5i) × (3 - 2i)3 - 2i= 23 - 2i + 6 - 4i + 15i - 10i2 3- 2i= 23 - 2i + 6 + 11i + 103 - 2i= 2 + 6 + 11i + 103 - 2i= 18 + 11i3 - 2i= 18 + 11i3 - 2i × 3 + 2i3 + 2i= 54 + 36i + 33i + 22i29 + 6i - 6i - 4i2= 54 + 69i -229 + 4z = 32 + 69i13z =32 + 69i132 = (32)2 + (69i)2 + 2(32)(69i)(13)2= 1024 - 4761 + 4416i169z = -3737 + 4416i169

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More