Question

Express the mentioned below complex number in the standard form $a+ib$
(vi) $\frac{(1+i)(1+\sqrt{3}i)}{1-i}$

Answer 1

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

$=\frac{1+\sqrt{3}i+i+\sqrt{3}{i}^2}{1-i}$

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

$=\frac{(1-\sqrt{3})+(\sqrt{3}+1)i}{1-i}\times \frac{1+i}{1+i}$

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

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

$=-\sqrt{3}+i$

$\therefore \frac{(1+i)(1+\sqrt{3}i)}{1-i}=-\sqrt{3}+i$