Question

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

Answer 1

`(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=(3i-39)/(2-i)`

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

`z=3((i-13)(2+i))/((2-i)(2+i))`

Multiply out,

`z=3(2i-1-26-13i)/(5)`

`z=3(-11i-27)/(5)`

The result is,

`z=-(81)/(5)-(33)/(5)i`

Hope this helps :)