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Suppose $f$ is an analytic function on a domain $D$. Then I need to show that $\overline {f(\bar z)}$ is also analytic. Here is what I did -

Suppose $f(z) = u(x,y) + iv(x,y)$ where $u$ and $v$ are real functions of $x$ and $y$ and $z = x + iy$. Now $f(\bar z) = u(x,-y) + iv(x,-y) $ and then $\overline {f(\bar z)} = u(x,-y) - iv(x,-y) $.

To show that a function is analytic, I need to verify that it satisfy Cauchy-Riemann Equations. I differentiated $\overline {f(\bar z)}$ and checked that it actually satisfies these equations. But here is my doubt - Being analytic means that the function if complex differentiable. Now here to check for analyticity, I am differentiating my function without proving that it's actually analytic. So is this the right way to do it?

Dark_Knight
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4 Answers4

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Alternatively, you can use that $z\mapsto \overline{z}$ is continuous to make: $$\begin{align} \lim_{h \to 0} \frac{\overline{f(\overline{z+h})}-\overline{f(\overline{z})}}{h} &= \lim_{h \to 0} \frac{\overline{f(\overline{z}+\overline{h})}-\overline{f(\overline{z})}}{\overline{\overline{h}}} \\ &= \overline{\lim_{h \to 0}\frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}}} \\ &= \overline{\lim_{\overline{h} \to 0}\frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}}} \\ &= \overline{\frac{\rm d}{{\rm d}z}\bigg|_{z=\overline{z}}f(z)}\end{align}$$

Ivo Terek
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A function is analytic if its real and imaginary parts (your $u$ and $v$) are differentiable, and those derivatives satisfy the Cauchy-Riemann equations. The $u$ and $v$ are real-valued functions of two real variables, so “differentiable” in this case just means differentiable as functions from $\Bbb{R}^2$ to $\Bbb{R}$; there's no further complex analysis to do.

So you just need to worry about whether the functions $(x,y) \mapsto u(x,-y)$ and $(x,y) \mapsto -v(x,-y)$ are differentiable for any real-differentiable functions $u$ and $v$. Which should be pretty clear, I think...

Micah
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Since $f$ is analytic, it can be represented as a power series, that is, for every $z_o\in D$,

$$f(z)=\sum_{n=0}^\infty c_n(z-z_0)^n$$

Let's assume that $z_0=0$. Then

$$\overline {f(\overline z)}=\sum_{n=0}^\infty c_n\overline {{z}}^n=\overline{ \lim_{k\to \infty}\sum_{n=0}^k c_n\overline{z}^n}$$ Since the function $z\mapsto \overline z $ is continuous,

$$\overline{ \lim_{k\to \infty}\sum_{n=0}^k c_n\overline{z}^n}=\lim_{k\to \infty}\overline{\sum_{n=0}^kc_n\overline{z}^n}= \lim_{k\to \infty}\sum_{n=0}^k \overline{c_n\overline{{z}^n}}=\lim_{k\to \infty}\sum_{n=0}^k\overline{c_n}z^n=\sum_{n=0}^{\infty}\overline{c_n}z^n$$ But

$$|c_nz^n|=|c_n||z^n|=|\overline{c_n}||z^n|=|\overline{c_n}z^n|$$

and then $\sum_{n=0}^{\infty}c_nz^n$ is absolutely convergent if and only if $\sum_{n=0}^\infty\overline{c_n}z^n$ is absolutely convergent. But $f(z)$ is analytic in $D$, so $\sum_{n=0}^{\infty}c_nz^n$ is absolutely convergent; then we can conclude that

$$\overline {f(\overline z)}=\sum_{n=0}^\infty \overline{c_n}{z}^n$$ is also analytic in $D$.

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When you differentiate a $C^1$ function $f(x, y)$ to check whether the CR equations hold, you are computing real (partial) derivatives; (complex) analyticity is the same as existence of the complex derivative. So yes, your method is correct.

Travis Willse
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