# Find the multiplicative inverse of the complex number $\;\normalsize 4-\normalsize 3 \normalsize i\;$

$(a)\; \normalsize \large\frac{2}{25}+\normalsize \large\frac{3}{25} \normalsize i\qquad(b)\; \normalsize \large\frac{4}{25}+\normalsize \large\frac{3}{25} \normalsize i\qquad(c)\; \normalsize \large\frac{3}{25}+\normalsize \large\frac{2}{25} \normalsize i\qquad(d)\; \normalsize \large\frac{4}{25}+\normalsize \large\frac{2}{25} \normalsize i$

Answer : $\;\normalsize \large\frac{4}{25}+\normalsize \large\frac{3}{25} \normalsize i$
Explanation :
Let $\;z=\normalsize 4-\normalsize 3 \normalsize i\;$
Then , $\;\overline{z} =\;\normalsize 4-\normalsize 3 \normalsize i\;and \; |z|^{2} = 4^{2} +(-3)^{2}$
$=16+9=25$
Therefore , the multiplicative inverse of $\;\normalsize 4-\normalsize 3 \normalsize i\;$ is given by
$z^{-1} =\large\frac{\overline{z}}{|z|^{2}} =\large\frac{\normalsize 4-\normalsize 3 \normalsize i}{\normalsize 25}$
$= \normalsize \large\frac{4}{25}+\normalsize \large\frac{3}{25} \normalsize i$
answered Apr 10, 2014 by