Re: An Easy Complex Quesition

"Greg Doyle" <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
> This should be an easy question, but I can't seem to get it.
> 
> Q: Solve the following equations in polar form and locate the roots in the
> complex plane:
> c) z^4 = -1 + \sqrt(-3)

What does sqrt(-3) mean? Is that i*sqrt(3) or does it
refer to +-i*sqrt(3)? I'll assume it means i*sqrt(3).

Then the magnitude of z^4 is 2 and the phase is 2*pi/3, i.e.

z^4 = 2*exp(i*2*pi/3)

> 
> This is taken from Bak, Newman, Complex Analysis. Pg 18.
> 
> They don't give the answer for the roots in the book, but clearly,
> (z^4)/2 = (-1 + \sqrt(-3))/2  is the third root of unity.
> Let w = (-1 + \sqrt(-3))/2
> Then when z = (2^{1/4}) w, z^4 = 2w. Done.
> 
> However, I'm stumped as to how to find the polar coordinates.
> 
> Can I just say
> z = 2^{1/4} e^{ (2 (PI) i )/ 3}
> Then r = 2^{1/4} and arg z = 2 (PI) i / 3, i = 1, 2

No. Review how you find the three units of unity:

1 = exp(i*(0 + 2*pi*k)), k = 0, 1, 2, ...

That is, you can add arbitrary multiples of 2*pi to the phase.

So 1^(1/3) = exp(i*(0+2*pi*k)/3)

k = 0 -> 1^(1/3) = 1
k = 1 -> 1^(1/3) = exp(i*2*pi/3)
k = 2 -> 1^(1/3) = exp(i*4*pi/3) = exp(-i*2*pi/3).

Now apply the same technique to the fourth root
of the above expression.

z^4 = 2*exp(i*2*pi/3)
     = 2*exp(i*[(2*pi/3) + 2*pi*k]), k = 0, 1, 2, ...

z = 2^(1/4)*exp(i*[(pi/6) + k*pi/2]), k = 0, 1, 2, ...

          - Randy