Question

11z/(z+3) = 13 - i , z∈C please solve it

Answer 1

$\frac{11z}{z+3}=13-i,z\in\mathbb{C}$

First, multiply both sides by $z+3$,

$11z=(13-i)(z+3)$

$13z+39-iz-3i=11z$

Collect terms and put z on the left,

$2z-iz=3i-39$

$z(2-i)=3i-39$

$z=\frac{3i-39}{2-i}$

Division of complex numbers is defined by multiplying both denominator and numerator with complex conjugate of the denominator,

$z=3\frac{(i-13)(2+i)}{(2-i)(2+i)}$

Multiply out,

$z=3\frac{2i-1-26-13i}{5}$

$z=3\frac{-11i-27}{5}$

The result is,

$z=-\frac{81}{5}-\frac{33}{5}i$

Hope this helps :)