Power Series expansion of Complex Functions

Question

Let $f :\Bbb C \to \Bbb C$ be the function, $f(z) = \overline z$ for all $z$. Prove that there is no $R > 0$ such that $f $ has a power series expansion on $D(0, R)$.

I have proved that the function is continuous on $\Bbb C$, but I don't know where to go from here?

Answer 1

You can check the Cauchy-Riemann equations. In general if $u + i v = f(z)$, $z = x + iy$, then the conditions are

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad\mbox{and}\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $$

In your case $u + iv = \bar{z} = x - iy$, so that

$$ \frac{\partial u}{\partial x} = 1 $$

whereas

$$ \frac{\partial v}{\partial y} = -1 $$

Clearly

$$ \frac{\partial u}{\partial x} \ne \frac{\partial v}{\partial y} $$

and then the function $f(z) = \bar{z}$ is not holomorphic and cannot be represented as a power series of $z$

Answer 2

Observe that, for $\;z=x+iy\;$ , we have that $\;f(z)=x-iy\;$, so the real part of $\;f\;$ is $\;u(x,y):=x\;$ , and its imaginary part is $\;v(x,y):=-y\;$ . Now check the Cauchy-Riemann equations of this thing:

$$u_x=1\neq0=v_y\;\;,\;\;u_y=0=-v_x$$

an thus the function is analytic at no point at all, so the function cannot have a power series around any point with positive convergence radius.

Answer 3

Without Cauchy - Riemann , holomorphy , etc...:

Suppose that $\overline{z}= \sum_{n=0}^{\infty}a_nz^n$ for $|z|<R$

$z=0$ gives $a_0=0$, thus

$\frac{\overline{z}}{z}=a_1+a_2z+a_3z^2+...$ for $0<|z|<R$.

It follows that

$(*)$ $\lim_{z \to 0} \frac{\overline{z}}{z}=a_1$.

If $z \in \mathbb R, z \ne 0$, $(*)$ gives $a_1=1$ and if $z \in i \mathbb R, z \ne 0$, $(*)$ gives $a_1=-1$ , a contradiction