# Thread: Sqaure roots of complex number

1. ## Sqaure roots of complex number

Hi,

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

$\sqrt{2}e^{(\frac{1}{2} + k)\pi i}$
$= \pm \sqrt{2}(cos(\frac{\pi}{12}) + isin(\frac{\pi}{12}))$

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 $arg = \frac{\pi}{6}, \frac{7\pi}{6}$

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

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 $z^2$
Then solve for Z....

3. Simply factor out a sqrt(2) from

4. ## Turns out like this for me

: $sqrt(3) +i$
$abs:z = 2$
$argz = atan1/sqrt3) = 30deg" alt="argz = atan1/sqrt3) = 30deg" />

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

$z = sqrt(2)(cos(pi/12) +isin(pi/12))$ + 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 :

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

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

6. ## Divide the argument

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

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

$
(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