Answer
$\dfrac{9}{5}+\dfrac{18}{5}i$
Work Step by Step
Multiplying by the conjugate of the denominator of $
\dfrac{9}{1-2i}
$ results to
\begin{array}{l}
\dfrac{9}{1-2i}
\cdot
\dfrac{1+2i}{1+2i}
\\\\=
\dfrac{9+18i}{(1)^2-(2i)^2}
\\\\=
\dfrac{9+18i}{1-4i^2}
\\\\=
\dfrac{9+18i}{1-4(-1)}
\\\\=
\dfrac{9+18i}{5}
\\\\=
\dfrac{9}{5}+\dfrac{18}{5}i
.\end{array}
