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Find the multiplicative inverse of $\;2 - 3 i$

$(a)\;\large\frac{2}{13}+ i \large\frac{3}{13}\qquad(b)\;\large\frac{2}{13}+ i \large\frac{5}{13}\qquad(c)\;\large\frac{5}{13}+ i \large\frac{7}{13}\qquad(d)\;\large\frac{3}{13}+ i \large\frac{5}{13}$

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Answer : $\;\large\frac{2}{13}+ i \large\frac{3}{13}$
Explanation :
Let $\;z=2-3i\;$ then $\;\overline{z} = 2+3i$
$|z|^{2} = (2)^{2}+(-3)^{2} 13$
Therefore , the multiplicative inverse of $\;2 - 3 i$ is given by
$z^{-1} = \large\frac{\overline{z}}{|z|^{2}} = \large\frac{2+3i}{13} = \large\frac{2}{13}+ i \large\frac{3}{13}$
The above working can be reproduced in the following manner also ,
$z^{-1} = \large\frac{2+3i}{(2-3i)(2+3i)}$
$= \large\frac{2+3i}{2^{2} - (3i)^{2}}$
$= \large\frac{2+3i}{13}$
$=\large\frac{2}{13}+ i \large\frac{3}{13}$
answered Apr 16, 2014 by yamini.v
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