# Complex Conjugates: f*(z) = f(z*)

## Main Question or Discussion Point

It's commonly known that if f(z) is analytic, then

f(z*) = f*(z)

that is, an analytic function of the complex conjugate is equal to the complex conjugate of the function...with the proviso that f(x+i0) = f(x) = Re f(x)

I've tried to prove it using the C-R equations but I'm not having much luck. Can anyone point me in the right direction?

Thanks.

Hurkyl
Staff Emeritus
Gold Member
If a function is analytic, then near each point in its domain, it's equal to its Taylor series (which has a positive radius of convergence)

This is known as the Reflection Principle in textbooks.

The two functions $$f(x)$$ and $$g(x) := \overline{f(\overline{x})}$$ are both analytic and they agree on the real line, which is a set with a limit point, therefore they agree everywhere.

Excellent.

Should have spotted the method where you write it as a Taylor series (necessarily with real coefficients, I guess, since the function's restriction to the real line must be real, as stated above) then apply complex conjugation to this series - then it just boils down to showing that the complex conjugate of a complex variable raised to a power is the power of the complex congugate of the variable. In other words

$$(z^n)^{*} = (z^{*})^n$$

which isn't too hard, I hope!

The Identity Theorem method is a bit more sublte, but it makes sense.

Thanks Hurkyl, g_edgar.

nrqed
Homework Helper
Gold Member
Excellent.

Should have spotted the method where you write it as a Taylor series (necessarily with real coefficients, I guess, since the function's restriction to the real line must be real, as stated above) then apply complex conjugation to this series - then it just boils down to showing that the complex conjugate of a complex variable raised to a power is the power of the complex congugate of the variable. In other words

$$(z^n)^{*} = (z^{*})^n$$

which isn't too hard, I hope!
Note that this is actually trivial if you use the polar form.

Note that this is actually trivial if you use the polar form.
Yep, was just thinking that.