# find a square root of the complex number

• Dec 28th 2010, 05:15 PM
Volga
find a square root of the complex number
Question. Find a square root of the complex number $\displaystyle i$.

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

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

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

The textbook answer, however, only gives one value of w ($\displaystyle w_1$). 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.
• Dec 28th 2010, 05:45 PM
pickslides
You are correct to say there is two solutions here $\displaystyle \displaystyle\pm\left(\frac{1}{\sqrt{2}}+\frac{i}{ \sqrt{2}}\right)$

Your book has given 'a' root, not all.
• Dec 28th 2010, 05:52 PM
Volga
Thanks for clarification, much appreciated!