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

$\dfrac{13z}{z+1} = 11-3i$

Observe that for $z\neq-1$,

$\dfrac{13z}{z+1} = \dfrac{13(z+1) - 13}{z+1} = 13 - \dfrac{13}{z+1}$

so we can simplify the equation to

$13 - \dfrac{13}{z+1} = 11 - 3i \implies \dfrac{13}{z+1} = 13 - (11 - 3i) = 2+3i$

Multiply both sides by $\frac{z+1}{2+3i}$.

$\dfrac{13}{z+1}=2+3i \implies \dfrac{13}{2+3i} = z+1 \implies z = \dfrac{13}{2+3i} - 1$

Rationalize the denominator by introducing its complex conjugate.

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