Question

The mulplicative inverse of complex number z =√3+2i
(a)z- is
(b)|z| is
(c) z-1. is
​

Answer 1

Given ,

The complex number is z =√3+2i

We know that , the mulplicative inverse of complex number z = a + ib is given by

$\boxed{ \tt{ {z}^{ - 1} = \frac{1}{z} = \frac{ \bar{z}}{ { |z| }^{2} } }}$

Where ,

$\tt \bar{z} =$ Conjugate of z

$\tt |z| =$ Modulus of z

Thus ,

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i }{ {\{ \sqrt{ {(3)}^{2} + {(2)}^{2} } \}}^{2} }$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i }{ { \{ \sqrt{3 + 4} \}}^{2} }$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i}{7}$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} }{7} - \frac{2i}{7}$

or

$\tt \implies {z}^{ - 1} = \frac{1}{ \sqrt{3} + 2i}$

$\tt \implies {z}^{ - 1} = \frac{1 \times ( \sqrt{3} - 2i)}{( \sqrt{3} +2i)( \sqrt{3} - 2i) }$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i }{ {( \sqrt{3} )}^{2} -{(2i)}^{2} }$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i }{ 3 - 4( - 1)}$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} - 2i }{ 7}$

$\tt \implies {z}^{ - 1} = \frac{ \sqrt{3} }{7} - \frac{2i}{7}$

Therefore , the multiplicative inverse is √3/7 - 2i/7

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