Question

Express $\frac{i(4+3i)}{(1-i)}$ in the form $a+ib$, where $a,b\in R$.

Answer 1

Now,

$\frac{i(4+3i)}{(1-i)}$

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

$=\frac{i(1+7i)}{2}$

$=\frac{(-7+i)}{2}$

$=-\frac{7}{2}+i\frac{1}{2}$

This is the required $a+ib$ form.