Simplify each of the following and express it in the form (a + ib) :
$(-2+\sqrt{-3})^{-1}$
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
$=\frac{-2-i \sqrt{3}}{(-2)^{2}-(i \sqrt{3})^{2}}$
$=\frac{-2-i \sqrt{3}}{4-\left(3 i^{2}\right)}$
$=\frac{-2-i \sqrt{3}}{4-3(-1)}\left[\because \mathrm{i}^{2}=-1\right]$
$=\frac{-2-i \sqrt{3}}{4+3}$
$=\frac{-2-i \sqrt{3}}{7}$
$=-\frac{2+i \sqrt{3}}{7}$
$=-\frac{2}{7}-\frac{\sqrt{3}}{7} i$
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