# Complex equations

1. Feb 9, 2009

### praharmitra

If i am solving for a complex equation f(z) = 0. can i assume that if i solve [f(z)]* = 0, I'll get the solution to the first one??

I mean, does f(z) = 0 imply [f(z)]* = f(z*) = 0???

I think it should be, but its giving me vague answers for a question i'm trying to solve. pls help

2. Feb 9, 2009

### Citan Uzuki

Well, obviously f(z)* = 0 if and only if f(z) = 0, since 0* = 0. On the other hand, it is not necessarily true that f(z*) = 0. For instance, if f(z) = i+z, then f(i*) = f(-i) = 0, but f(i) = 2i ≠ 0.

Now, it may be the case that you can prove that f(z*) = f(z)* for every z -- for instance, if f was a polynomial in z with real coefficients, this equation would hold, and then solving for f(z*) = 0 would get you a solution to the original equation. But it's not true in general.

3. Feb 10, 2009

### praharmitra

ya...i had realised that....but the question i am trying to solve, has real coefficients....and also...

[exp(z)]* = exp(z*) right????

I dont know why i am getting stuck.... neway. i'll post it in the homework section if i cant do it

4. Feb 10, 2009

### Citan Uzuki

Yes, that is correct.

5. Feb 10, 2009

### deiki

Yes, you can expand the exp function into powerseries. Plugging the properties a*+b*=(a+b)* and a*∙b*=(a∙b)* into these powerseries will lead you pretty quickly to [exp(z)]* = exp(z*)

This work for any holomorphic function you can expand into powerseries with real coefficients.

6. Feb 10, 2009

### praharmitra

ok... I found my mistake. I can't do what I have been trying to do. Turns our there's an imaginary coefficient after all.

Thanks anyway.