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Simplify 4+5i/4-5i

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  • Last Updated : 19 Mar, 2022

A complex number is a term that can be shown as the sum of real and imaginary numbers. 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 the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z) in form of a complex number.  It is also called an imaginary number. 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,

5 + 16i is a complex number, where 5 is a real number (Re) and 16i is an imaginary number (Im).

8 + 20i is a complex number where 8 is a real number (Re) and  20i is an imaginary number (Im)

Simplify 4+5i/4-5i 

Solution: 

Given: 4+5i/4-5i 

to simplifying multiply the numerator and denominator by the conjugate of denominator 

= (4+5i/4-5i) × (4+5i)(4+5i)

= {(4+5i)2}/ {(4)2– (5i)2}

= {16 + (5i)2 + 2(4)(5i)} / {16 – 25(i)2}

= {16 + 25(i)2 + 40i} / {16 + 25}

=  {16 – 25 + 40i} / 41

= (-9 + 40i) / 41

= -9/41 + 40/41i 

Similar Questions

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

Solution: 

Given : {(3-2i)(2+3i)} / {(1+2i)(2-i)}

                  = (6 +9i -4i – 6i2 ) / { 2 – i + 4i – 2i2 }

                  = (6+5i +6) / (2 – i +4i +2)

                  = (12+5i) /(4+3i)

Now use conjugate 

                  = {(12+5i) /(4+3i)} × { (4-3i)/(4-3i)}

                  = {(12+5i) ×(4-3i)} / { (4+3i)(4-3i)}

                  = {48 -36i +20i -15i2} / {(16 -(3i)2}

                  = {(48 -16i +15)}/ {(16 +9)}

                  = (63 -16i) / 25

                  = 63/25 -16/25i

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

Solution: 

Given : {(5 -2i)(5+3i)} / {(2+2i)(2-i)}

                 = (25 +15i -10i – 6i2 ) / { 4 – 2i + 4i – 2i2 }

                 = (25+5i +6) / ( 4 -2i +4i +2)

                 = (31+5i) /(6+2i)

now use conjugate

                 = {(31+5i) /(6+2i)} × {(6-2i)/(6-2i)}

                 = {(31+5i) ×(6-2i)} / {(6+2i)(6-2i)}

                 = {186 – 62i +30i -10i2} / {(36 -(4i)2}

                 = {(186 -62i +30i +10)}/ {(36 +4)}

                 = (196 – 32i) / 40

                 = 196/40 – 32/40i

                 = 49/10 – 8/10i

Question 3: Simplify (4 -2i)3?

Solution: 

Given : ( 4 -2i)3

Here we will use identity (a-b)3 =  a3 – b3 – 3a2b + 3ab2

           = (4)3 – (2i)3 – 3(4)2(2i) + 3 (4) (2i)2

           = 64 – (8i3) – 96i + 24i2

           = 64 -[8(-i)] – 96 i -24

           = 64 +8i – 96i – 24

           = 40 – 88i 

Question 4: Express the (2-8i) – (-5+6i)  in a+ib form?

Solution:

Given : (2 – 8i) – (-5+6i)

            = 2 – 8i + 5 -6i

            = 7 -14i

Question 5: Find the multiplicative inverse of -5 + 8i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as: 1 / z  or z-1 (Inverse of z)

        here z = -5 +8i

Therefore z = 1/z

                   = 1 / (-5 +8i)

Now rationalizing

                  = 1/(-5+8i) x (-5 – 8i)/(-5 -8i)

                  = (-5-8i) / {(-5)2 – 82i2}

                  = (-5 -8i) / {25 +64}

                  = (-5-8i)/ (89)

                  = -5/89 – 8i/89

Question 6: Simplify (5-7i)(7-8i)?

Solution: 

Given: (5-7i)(7-8i)

        = 35 – 40i – 49i + 56 i2

        = 35 -89i -56

        = -21 – 89i 

Question 7: Find the square of 2i /16?

Solution:

it gives a result in negative after squaring an imaginary number,  …

{(2/16)i}2 = (2/16 i) x (2/16 i)

               = 4/256 i2 

               = -4/256

               = – 1/64 i

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