# Thread: Sqaure roots of complex number

1. ## Sqaure roots of complex number

Hi,

The question i have is "Find the sqaure roots of $\displaystyle \sqrt{3} + i$ "

The book gives the answer:
$\displaystyle \sqrt{2}e^{(\frac{1}{2} + k)\pi i}$
$\displaystyle = \pm \sqrt{2}(cos(\frac{\pi}{12}) + isin(\frac{\pi}{12}))$

I can see how they get that but I thought you have to plug in k = 0, 1(n-1) and then use the the new arguements given from that to find them.

I get $\displaystyle arg = \frac{\pi}{6}, \frac{7\pi}{6}$

$\displaystyle \pm(\frac{\sqrt{6}}{2} + i\frac{\sqrt{2}}{2})$

Could anyone clear up my grave error?

Many thanks

2. ## Maybe

Im not sure of this will work but use Moivre´s.
Wright the complex number in polar form and set that to be $\displaystyle z^2$
Then solve for Z....
Hope it was helpful..

3. Simply factor out a sqrt(2) from

4. ## Turns out like this for me

:$\displaystyle sqrt(3) +i$
$\displaystyle abs:z = 2$
$\displaystyle argz = atan1/sqrt3) = 30deg$

$\displaystyle => 2(cos(pi/6)+isin(pi/6)$
Set that to be $\displaystyle z^2$

$\displaystyle z = sqrt(2)(cos(pi/12) +isin(pi/12))$ + the other root that gave the +/- sign to the equation...

5. Thanks for the tips,

When I take out the factor of root 2 I have :

$\displaystyle \pm \sqrt{2}(cos(\frac{\pi}{6}) +isin(\frac{\pi}{6}))$

Where apparently im supposed to have $\displaystyle \theta = \frac{\pi}{12}$

6. ## Divide the argument

Need to devide the argument (pi/6) by 2. $\displaystyle Z^n$ has $\displaystyle n$ roots, the arguments of the roots are evently spread around a circle.

To find roots of $\displaystyle z^2$ devide the argument by $\displaystyle 2 => (pi/6)2$ and that gives you pi/12.... the other roots are found by:

$\displaystyle (2*pi*n)/2 = pi*n$ where your n can be 1;2

7. Perhaps the following attachment may help

8. Great, thanks for all the help.

As a side note could anyone recommend a book/website with exercises to do with complex number? I have 2 already but they lack exercises.

Thanks