#### Answer

$\dfrac{9}{1-2i}=\dfrac{9}{5}+\dfrac{18}{5}i$

#### Work Step by Step

$\dfrac{9}{1-2i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{9}{1-2i}=\dfrac{9}{1-2i}\cdot\dfrac{1+2i}{1+2i}=\dfrac{9(1+2i)}{1^{2}-(2i)^2}=\dfrac{9(1+2i)}{1-4i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{9+18i}{1-4(-1)}=\dfrac{9+18i}{1+4}=\dfrac{9+18i}{5}=\dfrac{9}{5}+\dfrac{18}{5}i$