Question

Solve the equation 13Z/Z+1 =11-3i, Z£C where Z=X+yi X&Y£R?​

Answer 1

Presumably you mean the equation

`(13z)/(z+1) = 11-3i`

Observe that for
`z\\eq-1`,

`(13z)/(z+1) = (13(z+1) - 13)/(z+1) = 13 - (13)/(z+1)`

so we can simplify the equation to

`13 - (13)/(z+1) = 11 - 3i \implies (13)/(z+1) = 13 - (11 - 3i) = 2+3i`

Multiply both sides by
`(z+1)/(2+3i)`.

`(13)/(z+1)=2+3i \implies (13)/(2+3i) = z+1 \implies z = (13)/(2+3i) - 1`

Rationalize the denominator by introducing its complex conjugate.

`z = (13(2-3i))/((2+3i)(2-3i)) - 1 = (26-39i)/(2^2+3^2) - 1 = \boxed{1 - 3i}`