Question

If $a+ib=\frac{-1+\sqrt{3}i}{1+i}$, then find $a$ and $b$.

Answer 1

We have,

$a+ib=\frac{-1+\sqrt{3}i}{1+i}$

$a+ib=\frac{-1+\sqrt{3}i}{1+i}\times \frac{1-i}{1-i}$

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

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

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

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

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

On comparing both sides, we get

$a=\frac{(\sqrt{3}-1)}{2},b=\frac{(\sqrt{3}+1)}{2}$

Hence, this is the answer.