Answer
$-\dfrac{3}{2}+\dfrac{5}{2}i$
Work Step by Step
Multiplying by the conjugate of the denominator and using $i^2=-1,$ the given expression, $
\dfrac{1+4i}{1-i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1+4i}{1-i}\cdot\dfrac{1+i}{1+i}
\\\\=
\dfrac{1(1)+1(i)+4i(1)+4i(i)}{1^2-i^2}
\\\\=
\dfrac{1+i+4i+4i^2}{1-i^2}
\\\\=
\dfrac{1+5i+4(-1)}{1-(-1)}
\\\\=
\dfrac{1+5i-4}{1+1}
\\\\=
\dfrac{-3+5i}{2}
\\\\=
-\dfrac{3}{2}+\dfrac{5}{2}i
.\end{array}
