Raising a complex number to the nth power

1. Jul 16, 2008

magda3227

I was looking around a little bit for an algorithm that would compute a complex number to the nth power.

Can anyone supply me a resource that covers this? I wouldn't imagine it being different than some sort of (x+y)^n formula.

2. Jul 16, 2008

yenchin

You can convert the complex number into either polar form or exponential form and then use De Moivre's theorem.

3. Jul 16, 2008

HallsofIvy

Staff Emeritus
DeMoivre's theorem:
If $z= r e{i\theta}$, then $z^n= r^ne^{i n\theta}$.

r is the "modulus" or absolute value of z: if z= x+iy then $|z|= \sqrt{x^2+ y^2}$.

$\theta$ is the "argument" or angle the line through 0 and z in the complex plane makes with the real-axis: if z= x+ iy, then $\theta= arctan(y/x)$.

4. Jul 16, 2008

maze

@Magda:
Here's a question for you to ponder: if you take a complex number, z, that has length 1 (a^2 + b^2 = 1 for z = a+bi), then what happens when you keep raising it to higher and higher powers: z^1, z^2, z^3, ... z^100000, ...

5. Jul 16, 2008

magda3227

Thank you all very much. I was not familiar with DeMoivre's Theorem at all. I have seen Euler's identity, however.

In response to maze, I have no idea what happens when you raise a complex number, z = length 1 to higher and higher powers. I can't even begin to make an assumption of what would happen.

I'm not smart. :/

6. Jul 16, 2008

maze

Try some examples!

Here are the most obvious ones:
1 1 1 1 1 1 1 1 ...
i -1 -i 1 i -1 ...

Here are some for you to try:
1/sqrt(2) + i/sqrt(2)
-1/2 + i*sqrt(3)/2

7. Jul 17, 2008

HallsofIvy

Staff Emeritus
It's not a matter of being smart, it's a matter of having specific knowledge. It is true, generally, that |xn|= |x|n. In particular, if |z|= 1 then every power of z will also have absolute value 1. In the complex plane, the absolute value of a number is its distance from the origin. Every number with absolute value 1 lies on the unit circle. If z is on the unit circle, the so is zn for all n, although they may move around the unit circle.

8. Jul 17, 2008

maze

was hoping he would figure this out on his own....