Answer
$\dfrac{6i}{1-2i}=-\dfrac{12}{5}+\dfrac{6}{5}i$
Work Step by Step
$\dfrac{6i}{1-2i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{6i}{1-2i}=\dfrac{6i}{1-2i}\cdot\dfrac{1+2i}{1+2i}=\dfrac{6i(1+2i)}{1^{2}-(2i)^{2}}=\dfrac{6i+12i^{2}}{1-4i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{6i+12(-1)}{1-4(-1)}=\dfrac{-12+6i}{1+4}=\dfrac{-12+6i}{5}=-\dfrac{12}{5}+\dfrac{6}{5}i$
