# Sqaure roots of complex number

• Jan 14th 2010, 07:57 AM
aceband
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
• Jan 14th 2010, 08:06 AM
Henryt999
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....
• Jan 14th 2010, 08:13 AM
Calculus26
• Jan 14th 2010, 08:15 AM
Henryt999
Turns out like this for me
: $sqrt(3) +i$
$abs:z = 2$
$argz = atan:(1/sqrt3) = 30deg$

$=> 2(cos(pi/6)+isin(pi/6)$
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...
• Jan 14th 2010, 08:19 AM
aceband
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 $\theta = \frac{\pi}{12}$
• Jan 14th 2010, 08:40 AM
Henryt999
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
• Jan 14th 2010, 08:58 AM
Calculus26
Perhaps the following attachment may help
• Jan 14th 2010, 02:18 PM
aceband
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