Find square root of $- 4 - 3 i$ .

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Solution

Now,
$- 4 - 3 i$
$= \frac{1}{2} (- 8 - 6 i)$
$= \frac{1}{2} {(3 i)^{2} + (1)^{2} - 2.3 i .1}$ [Since $i^{2} = - 1$ ]
$= \frac{1}{2} {(3 i - 1)^{2}}$
Now square root of $- 4 - 3 i$ is $\pm \frac{1}{\sqrt{2}} (3 i - 1)$ .

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