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I took a Complex Analysis Exam a few days ago and my grade unfortunately wasn't to expectation. I lost most of my points to the following question, but I'm still not sure how to prove it.

"Let $f: \mathbb{C}\to \mathbb{C}$ be an analytic function.

Prove whether $ g(z) = \overline{f(\bar{z})}^2$ is analytic."

What I initially did was state that $f(z) = z$, but I'm under the impression that was my mistake; though, I'm honestly not sure. As far as what I did after, I plugged it into the formula to find $\bar{f(\bar{z})}^2 = \bar{\bar{z}^2} = $, which I eventually broke into the components $\bar{\bar{z}^2} = z^2 = x^2-y^2+i2xy$. I then showed that the Cauchy-Riemann Equations were satisfied at every point, therefore the function was analytic.

My Professor crossed off what I wrote after the $z^2$, but he didn't write what was wrong; so I'm at a loss of what to do. Any help would be appreciated.

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Your mistake was to state that $f(z)=z$, since there are other entire functions. For proof of analytic it's, consider proving first that $\overline{f(\bar z)}$ is analytic by looking at Taylor series. – studiosus Apr 29 '14 at 2:40
You may check by using C-R equation. Let $z = x + iy$ and $f(z) = u(x,y)+iv(x,y)$, then $\overline{f(\bar{z})} = u(x,-y) - i v(x,-y)$. Now you are left to do some elementary derivatives. – zeno tsang Apr 29 '14 at 2:53
Thank you for your insight. I haven't learnt of using Taylor Series yet in class, but I now understand why I can't simply assign z equal to the function. Having two methods to prove it as analytic makes it that much easier for me to understand though; thank you. – Daniel Apr 29 '14 at 3:25
up vote 1 down vote accepted

You've correctly identified your problem: The fact that $f$ is analytic doesn't imply that $f(z) = z$ (certainly $e^z$, $z^2$, $\cos z$ are analytic). Of course, if you were coming up with a counterexample, it would be perfectly fine to let $f(z) = z$ (or whatever the counterexample should be) and prove the statement false. However, the statement actually is true, so this approach is hopeless.

One way to do this is to use the fact that $f$ is analytic if it's equal to its Taylor series. Now using the continuity of the operation of complex conjugation is analytic (together with the fact that $\overline z^k = \overline{z^k}$), so if

$$f(z) = \sum_{k = 0}^{\infty} a_k z^k$$

we have

\begin{align*} \overline{f({\overline z})} &= \overline{\sum_{k = 0}^{\infty} a_k \overline z^k} \\ &= \sum_{k = 0}^{\infty} \overline{a_k \overline z^k} \\ &= \sum_{k = 0}^{\infty} \overline{a_k} z^k \end{align*}

Now since $|a_k| = |\overline{a_k}|$, both series have the same (infinite) radius of convergence, so $\overline{f(\overline z)}$ is simultaneously analytic with $f$. Now the square of an analytic function is analytic, and we're done.

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Thank you very much again for the help. I'm not sure if my Professor was looking for us to use a Taylor Series in particular, but I'm understanding the answer better now. I originally used z to just represent analytic function, but I realise I hadn't thought it through properly now. To verify, any analytic function can be represented by an infinite power series? We haven't covered Taylor Series for Complex Variables, so I'm unfamiliar with them in a non-real context if they're different at all. – Daniel Apr 29 '14 at 3:21

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