The multiplicative inverse of (3+2 i)^2 is - Vidyadip

MCQ

The multiplicative inverse of $(3+2 i)^2$ is

A
$\left(\frac{-4}{169}+\frac{12}{169} i\right)$
✓
$\left(\frac{5}{169}-\frac{12}{169} i\right)$
C
$\left(\frac{-5}{169}+\frac{12}{169} i\right)$
D
$\left(\frac{5}{169}+\frac{12}{169} i\right)$

✓

Answer

Correct option: B.

$\left(\frac{5}{169}-\frac{12}{169} i\right)$

$z=(3+2 i)^2=\left(9+4 i^2+12 i\right)$
$=(9-4+12 i)=(5+120)$
$\Rightarrow z^{-1}=\frac{1}{(5+12 i)} \times \frac{(5-12 i)}{(5-12 i)}=\frac{(5-12 i)}{\left(25-144 i^2\right)}=\frac{(5-12 i)}{(25+144)}=\frac{(5-12 i)}{(16 i)}$
$\Rightarrow z^{-1}=\left(\frac{5}{169}-\frac{12}{169} i\right)$

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