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# $z_{1}=4+5i,z_{2}=-3+2i$ then $\large\frac{z_{1}}{z_{2}}$is

$\begin{array}{1 1}(1)\large\frac{2}{13}-\frac{22}{13}i&(2)-\frac{2}{13}+\frac{22}{13}i\\(3)\frac{-2}{13}-\frac{23}{13}i&(4)\frac{2}{13}+\frac{22}{13}i\end{array}$

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A)
$z_1=4+5i$
$z_2=-3+2i$
$\large\frac{z_1}{z_2}=\frac{4+5i}{-3+2i}$
$\qquad= \large\frac{(4+5i)(-3-2i)}{(-3+2i)(-3-2i)}$
$\qquad= \large\frac{-12-15 i -8 i -10 i^2}{(-3)^2-(2i)^2}$
$\qquad= \large\frac{-12+10-23i}{9+4}$
$\qquad= \large\frac{-2}{13}-i \frac{23}{13}$
Hence 3 is the correct answer.