Presumably you mean the equation
[tex]\dfrac{13z}{z+1} = 11-3i[/tex]
Observe that for [tex]z\neq-1[/tex],
[tex]\dfrac{13z}{z+1} = \dfrac{13(z+1) - 13}{z+1} = 13 - \dfrac{13}{z+1}[/tex]
so we can simplify the equation to
[tex]13 - \dfrac{13}{z+1} = 11 - 3i \implies \dfrac{13}{z+1} = 13 - (11 - 3i) = 2+3i[/tex]
Multiply both sides by [tex]\frac{z+1}{2+3i}[/tex].
[tex]\dfrac{13}{z+1}=2+3i \implies \dfrac{13}{2+3i} = z+1 \implies z = \dfrac{13}{2+3i} - 1[/tex]
Rationalize the denominator by introducing its complex conjugate.
[tex]z = \dfrac{13(2-3i)}{(2+3i)(2-3i)} - 1 = \dfrac{26-39i}{2^2+3^2} - 1 = \boxed{1 - 3i}[/tex]
