find a square root of the complex number

**Volga** · December 28th 2010, 05:15 PM

Question. Find a square root of the complex number $i$ .

Answer.
Let $i^2=w$ . By inspection (of i), the modulus of w is 1; we need to find $\theta$ .

$(\cos(\frac{\theta}{n}+\frac{2kn}{\pi})+i\sin(\fra c{\theta}{n}+\frac{2kn}{\pi})$ where n=2 and k is 0 and 1 (from 0 to n-1).

Therefore there are two numbers w that, when squared, give $i$ :
$\theta_1=\frac{\pi}{4}$ and $\theta_2=\frac{5\pi}{4}$
$w_1=\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$
$w_2=-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$

The textbook answer, however, only gives one value of w ( $w_1$ ). Am I wrong in thinking that there are two values of w? Or, the question has only asked for one of them ('a square root'), rather than 'find all square roots'?

PS I understand that there are always two numbers (one positive and one negative) that, when squared, give the same number. I only wanted to check whether the same principle applies to the questions about complex numbers.

**pickslides** · December 28th 2010, 05:45 PM

You are correct to say there is two solutions here $\displaystyle\pm\left(\frac{1}{\sqrt{2}}+\frac{i}{ \sqrt{2}}\right)$

Your book has given 'a' root, not all.

**Volga** · December 28th 2010, 05:52 PM

Thanks for clarification, much appreciated!

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