Answer
$-\dfrac{1}{2}+\dfrac{5}{2}i$
Work Step by Step
Multiply the numerator and the denominator by the conjugate of $1-i$ which is $1+i$.
$\dfrac{2+3i}{1-i}=\dfrac{2+3i}{1-i}\cdot \dfrac{1+i}{1+i}$
Use FOIL method in the numerator and special formula $(a+b)(a-b)=a^2-b^2$ in the denominator.
$=\dfrac{2(1)+2(i)+3i(1)+3i(i)}{1^2-i^2}$
$=\dfrac{2+2i+3i+3i^2}{1-i^2}$
Use $i^2=-1$.
$=\dfrac{2+2i+3i+3(-1)}{1-(-1)}$
$=\dfrac{2+2i+3i-3}{1+1}$
Simplify.
$=\dfrac{-1+5i}{2}$
$=-\dfrac{1}{2}+\dfrac{5}{2}i$
Hence, the solution in the standard form is $-\dfrac{1}{2}+\dfrac{5}{2}i$.
