Entire functions problem

Question

Suppose that $f:\Bbb C\to \Bbb C$ is entire, and denote by $u,v$ two functions $\Bbb R^2 \to \Bbb R$ such that $$f(x+iy)=u(x,y)+iv(x,y),$$ $\forall x,y \in \Bbb R$

Part (a) Show that if u is constant, then f is constant.

My attempt If $u(x,y)$ is constant then $\frac{\partial u}{\partial x}=0$ and $\frac{\partial u}{\partial y}=0$ therefore the riemann cauchy equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=0$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ therefore $v(x,y)$ must also be constant which in turn means that f has to be constant, since $u(x,y),v(x,y)$ are constant Can anyone confirm if my approach for part (a) is correct?

Part (b)

Show that the function $z\mapsto \overline {f(\bar{z})}$ is entire.

My attempt First calculating what $\overline {f(\bar{z})}$ looks like, $f(\bar {z})=u(x,y)-iv(x,-y)$ therefore $$\overline {f(\bar{z})}=-u(x,y)+iv(x,-y)$$ i tried next to check if this satisfies the riemann cauchy equations but they dont so maybe i've made a mistake in my working of calculating $\overline {f(\bar{z})}$

Part (c)

Let $D(0;1):=\{(x,y)\in \Bbb R^2:x+iy\in D(0;1)\}$ denote the open unit disk. Consider the function $g:D(0;1)\to \Bbb C$ defined by $$g(z):=\frac{1}{1-z}$$ $\forall z \in \Bbb C$ , we also denote $$E:=\{(x,y)\in \Bbb R^2:x+iy\in D(0;1)\}$$

The question is to find the functions $u,v:E\to \Bbb R$ such that $g(x+iy)=u(x,y)+iv(x,y)$ $\forall (x,y)\in E$ My attempt: $$g(z)=\frac{1}{1-z}=\frac{1}{1-(x+iy)}=\frac{1}{1-x-iy}$$ I have gotten stuck here can i split this fraction up to calculate $u(x,y)$,$v(x,y)$. Then show that $u(x,y)$ and $v(x,y)$ satisfies the riemann cauchy equations? would this be the end of the question or is this step not needed as it just asks to calculate $u(x,y)$ and $v(x,y)$

If anyone could go through this and tell me if I've made mistakes and if so where I've made the mistakes i would be extremely thankful.

Answer 1

Your argument for (a) is sound. For (b), I think you just have a sign error. You should get $$ \overline{f(\overline{x+iy})} = u(x,-y) - i v(x,-y) $$ Try checking the Cauchy-Riemann equations from there. (EDIT: For (b), see the other answer, which is better.) For (c), try "rationalizing" the denominator by multiplying by the complex conjugate over itself. $$ \frac{1}{1-x-iy} \left(\frac{1-x+iy}{1-x+iy}\right) = \ldots $$

Answer 2

(a) is ok. For (b) you can do $$\lim_{h \to 0} \frac{\overline{f(\overline{z+h})} - \overline{f(\overline{z})}}{h} = \mbox{fill the gap} = \overline{f'(\overline{z})}.$$For (c), use $$\frac{1}{1-z} = \frac{1-\overline{z}}{1-2\,{\rm Re}(z) +|z|^2},$$and now write $z = x+iy$. The denominator is real so you shouldn't have any trouble now.