Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 03/Solution 1

We want to have an answer in the form $a+ib$. To get rid of the imaginary part of the denominator $1-i$, we multiply both the numerator and the denominator by the complex conjugate $1+i$ of the denominator.

${\frac {z}{u}}={\frac {1+3i}{1-i}}={\frac {1+3i}{1-i}}\cdot {\frac {1+i}{1+i}}={\frac {(1+3i)(1+i)}{(1-i)(1+i)}}.$

Then, by the hint, the denominator becomes a real number – namely, $|1-i|^{2}$. Hence

${\frac {(1+3i)(1+i)}{(1-i)(1+i)}}={\frac {1+i+3i-3}{1^{2}+1^{2}}}={\frac {-2+4i}{2}}=\color {blue}-1+2i.$