Simplify each of the following and express it in the form (a + ib) :

Given: $(-2+\sqrt{-3})^{-1}$

We can re- write the above equation as

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

$=\frac{1}{-2+\sqrt{3 i^{2}}}\left[\because i^{2}=-1\right]$

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

Now, rationalizing

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

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

Now, we know that,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

So, eq. (i) become

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

$=\frac{-2-i \sqrt{3}}{4-\left(3 i^{2}\right)}$

$=\frac{-2-i \sqrt{3}}{4-3(-1)}\left[\because \mathrm{i}^{2}=-1\right]$

$=\frac{-2-i \sqrt{3}}{4+3}$

$=\frac{-2-i \sqrt{3}}{7}$

$=-\frac{2+i \sqrt{3}}{7}$

$=-\frac{2}{7}-\frac{\sqrt{3}}{7} i$