# Complex analysis

1. Jan 24, 2009

### shannon

1. The problem statement, all variables and given/known data
Prove that (z̄ )^k =(z̄ ^k) for every integer k (provided z≠0 when k is negative)

2. Relevant equations

3. The attempt at a solution
I let z=a+bi so, z̄ =a-bi
Then I plugged that into one side of the equation to get
(a-bi)^k
I was going to try to manipulate this to get [(a-bi)^k]
But I don't know where to go from here, or even what to do.....

2. Jan 24, 2009

### nrqed

This is a bit unclear. From the rest of your post, it looks as if by ̄ you meant the complex conjugation. So did you mean to write

$$(z^*)^k = (z^{* k})$$ ??

But even this is unclear since most people would interpret the two sides to mean the same thing.

It would make more sense if the question was to prove

$$(z^*)^k = (z^k)^*$$

was that the question?

If so, then write the complex number z in polar form $r e ^{i \theta}$

3. Jan 24, 2009

### NoMoreExams

Is the bar over the z only in both parts? Or is it

$$\bar{z}^{k} = \bar{\left(z^k\right)}$$

4. Jan 24, 2009

### shannon

Yes, that is my question. Sorry for the lack of clarity.
So I rewrote z in polar form, but I wasn't sure about what the conjugate of that would be...would the exponent just be negated? (the iѲ term).
If that was the case, would I simply combine the exponents in the left side of the equation, multiplying the k and the iѲ term thus resulting in the right side of the equation?
I hope I'm being more clear!

5. Jan 24, 2009

### nrqed

No problem!
Yes. Complex conjugation just means that we replace any "i" we see by -i. So yes, the sign of the exponent changes.
Exactly! Basically, k times (i)* gives the same thing as (k times i)*