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How to Divide Complex Numbers?

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Complex number is the sum of a real number and an imaginary number. These are the numbers that can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z.

Here in complex number form the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z). In complex number form a +bi, ‘i’ is an imaginary number called “iota”. 

The value of i is (√-1) or we can write as i2 = -1.

For example:

  • 3+4i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
  • 2+5i is a complex number where 2 is a real number (Re) and  5i is an imaginary number (im)

The Combination of real number and imaginary number is called a Complex number.

Imaginary numbers

The numbers which are not real are termed Imaginary numbers. After squaring an imaginary number, it gives a result in negative. Imaginary numbers are represented as Im().     

Example: √-3, √-7, √-11 are all imaginary numbers. here ‘i’ is an imaginary number called “iota”.

How to divide complex numbers?

Solution: 

There are few steps to divide the complex number:

The process of dividing  two complex numbers is slightly different from that of the division process of two real numbers. It is the concept of rationalizing the denominator in the case of fractions involving irrational numbers as their denominators.  

The following steps are involved:

Step 1: Assure that both the numerator and denominator are in the standard form of complex numbers, i.e., z = a + ib.

Step 2: Calculate the conjugate of the complex number that is at the denominator of the fraction. Say, if the denominator is a + bi, then its conjugate is a − bi.

Step 3: Multiply the numerator and denominator with the conjugate or with both the terms of the fraction.

Step 4: Simplify the numerator with distributive property

Step 5: Simplify the denominator with the use of difference of squares formula . i.e (a+b)(a-b) = a2 – b2

Step 6: In last of complex number separate  its real and imaginary parts.

Suppose there are two complex numbers z1 = a+bi, z2 = m+ni

Divide: a+bi / m+ni = {(a+bi)(m-ni)} / {(m+ni)(m-ni)}

                                = {am – ani + mbi – bni2 } / (m2 + n2)

                                = {am – ani + mbi – bn (-1)} / (m2 + n2)

                                = {am – ani + mbi +bn} / (m2 + n2)

                                = {(am + bn) + (mb – an)i} / (m2+n2)

                                = {(am + bn) / (m2 +n2)} + {(mb-an)/ (m2 +n2)} i

here  {(am + bn) / (m2 +n2)} is real part and {(mb-an)/ (m2 +n2)} i is imaginary part

Sample Problems

Question 1: Simplify {(-3 – 5i) / (2 +2i)}?

Solution:

Given {(-3 – 5i) / (2 +2i) }

conjugate of denominator 2+2i is 2-2i

Multiply the numerator and denominator with the conjugate 

Therefore {(-3 – 5i) / (2 +2i) } x {(2-2i) / (2-2i) }

              = {-6 -6i -10i +10i2  } /  { 22 – (2i)2 }                  {difference of squares formula . i.e (a+b)(a-b) = a2 – b2 }

              =  { -6 -6i -10i + 10 (-1) } /  { 4 – 4(-1) }         { i2 = -1 }

              = {  -6 -6i -10i -10 } / { 4 + 4 } 

              = (-16 – 16i ) / 8 

              = -16 /8  – 16i /8

              = -2  -2i

Question 2: Simplify {(-3 + 5i) / (2 – 2i)}?

Solution: 

Given {(-3 + 5i) / (2 – 2i)}

conjugate of denominator 2-2i is 2+ 2i

Multiply the numerator and denominator with the conjugate

Therefore {(-3 + 5i) / (2 -2i) } x {(2+2i) / (2+2i) }

             = {-6 + 6i +10i -10i2  } /  { 22 – (2i)2 }                  {difference of squares formula . i.e (a+b)(a-b) = a2 – b2 }

             = {-6 + 6i +10i – 10 (-1) } /  { 4 – 4(-1)}               { i2 = -1 }

             = {-6 + 6i +10i +10 } / { 4 + 4 }

             = (4 +16i ) / 8

             = 4 /8  +16i /8

             = 1/2  + 2i

Question 3: Solve (1-5i) / (-3i)?

Solution:

Given : (1-5i) / (-3i)

standard form of denominator -3i = 0 – 3i

conjugate of denominator     0-3i = 0 +3i

Multiply the numerator and denominator with the conjugate

therefore, {(1-5i) / (0 -3i) } x { (0+3i )/( 0 +3i )}

              = { (1-5i)(0+3i )  } / { 0 – (3i)2 }

              = { 3i – 15i2 } / { 0 –  (9(-1) ) }

              = { 3i – 15 (-1) } / 9

              = ( 3i +15 ) / 9

              = 15 / 9 + 3i/ 9

              = 5/3  + (1/9) i

Question 4: Perform the following operation and find the result in form of a +ib?

(2 – √-25) / (1 – -16)

Solution: 

Given (2 – √-25 ) / (1 – √-16)                                                  

(2 – √-25 ) / (1 – √-16) =  {2 – (i)(5)} / { 1 – (i)(4) }                 { i = √-1 }

=   (2-5i ) / (1-4i)

= { (2-5i ) / (1-4i)} x {(1+ 4i) / (1+ 4i)}

=   { (2-5i ) (1+ 4i) } / {  (1-4i) (1+4i) }

=   { 2 +8i -5i -(20i2) } /  { (1-16i2)}                                 { i2 = -1 }

= { 2 +3i +20} / {1 – 16(-1) }

= (22 +3i ) / ( 1 +16)

= (22+3i)/17

= { (22/17) + (3i/17) }

= 22/17  + 3i/17

Question 5: If z1, z2 are (1-i), (-2 +4i), respectively, find Im(z1z2/z1

Solution: 

Given : z1 = (1-i) 

                            z2 = (-2 +4i)

now to find Im(z1z2/z1

put values of z1 and z2

        Im(z1z2/z1)     = {(1-i) (-2 +4i) } / (1-i)

                               = {( -2 +4) +i(2+4)} / (1+i)

                               = {( 2+6i) /(1+i)}

                               = {( 2+6i) /(1+i)} x { (1+i) / (1- i) }

                               = {( 2+6i) (1+i)} / {(1+i)(1-i)}

                               = {(2+6) +i(6-2) } / (1+1)

                               = 4+2i

therefore Im(z1z2/z1) = 2

Question 6: Express (1-i)/(1+i) in standard form and find their conjugate?

Solution: 

Given : (1-i)/(1+i)

= {(1-i)/(1+i) x (1-i)/(1-i)}

= { (1-i)2} / {(1)2 -(i)2}

= {1 -2i -1 } / (1-i2)

= -2i / 2

= 0 – 2i/2

= 0 -i

now conjugate =  0+i

Question 7: Express in form of a+ib, 3(7+7i) + i(7+7i)

Solution:   

Given : 3(7+7i) + i(7+7i)

                        = 21 + 21i + 7i + 7i2

                        = 21 + 28i  + 7(-1)

                        = 21 +28i -7

                        = 14 +28i



Last Updated : 09 Mar, 2022
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