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# If $(m-5)+i(n+4)$ is the complex conjugate of $(2m+3)+i(3n-2)$ then $(n ,m )$are

$\begin{array}{1 1}(1)\bigg(-\frac{1}{2},-8\bigg)&(2)\bigg(-\frac{1}{2},8\bigg)\\(3)\bigg(\frac{1}{2},-8\bigg)&(4)\bigg(\frac{1}{2},8\bigg)\end{array}$

Can you answer this question?

Given $(m-5) +i (+4) =(2m+3)+i(3-2)$
$(m-5) +i(n+4) =(2m+3) -i (3n-2)$
$m-5=2m+3$
$n+4=-(3-2)$
$m=8$
$4n=2-4$
$n= -\large\frac{2}{4}=-\large\frac{1}{2}$
$(n,m)=\bigg(-\large\frac{1}{2}$$,-8 \bigg)$
Hence 1 is the correct answer.
answered May 14, 2014 by