Question
Simplify the following and express in the form a + ib:
$\frac{4+3 i}{1-i}$
✓
Answer
$\begin{aligned} \frac{4+3 i}{1-i} & =\frac{(4+3 i)(1+i)}{(1-i)(1+i)} \\ & =\frac{4+4 i+3 i+3 i^2}{1-i^2} \\ & =\frac{4+7 i+3(-1)}{1-(-1)} \quad \ldots\left[\because i^2=-1\right]\end{aligned}$
$=\frac{1+7 i}{2}=\frac{1}{2}+\frac{7}{2} i$
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