Square root of complex number in rectangular form

[chebyshevF]

Homework Statement

I don't know how to find the square root of a complex number in rectangular form?
As in, say, [tex]\sqrt{}9-6i[/tex]..my calculator can't do such an operation (yet my graphics calculator can, which can't be used in exams), so how do i go about to do this 'by hand'?

I just found this site: http://mathworld.wolfram.com/SquareRoot.html half way down, is that the formula that we use? How do we take the inverse tangent of (x,y)? What's sgn? I've never been taught such a formula nor did I know something like it existed until now...this question is in regards to a third year electromagnetics course and some 3 questions on the tutorials involve square roots of complex numbers..yet I'm assuming the professor found them using a graphics calculator.

 

[jbunniii]

Perhaps the most straightforward way:

1. convert to polar coordinates ([itex]re^{i\theta}[/itex])
2. take the square root ([itex]\sqrt{r}e^{i\theta/2}[/itex])
3. convert back to rectangular coordinates ([itex]\sqrt{r}(\cos(\theta/2)+i\sin(\theta/2))[/itex]

This gives you one square root. There are two. What's the other one?

P.S. This is equivalent to equation (1) at the link you provided.

 

[chebyshevF]

^^I actually did that, and I get the magnitude part right but i don't know how to get the angle/imaginary part? Or do i just multiply it all out by the (cos(theta/2) + isin(theta/2)) part?

 

[chebyshevF]

Yep that's what you do! Just tried it out and it works, thanks for pointing me in the right direction :)