# Analytic complex function which is constant.

I come across these question when I am studying George Cain Complex analysis.

1. Suppose $f$ is analytic on a connected open set $D$, and suppose $f^{'}(z)=0$ for all $z\in D$. Prove that $f$ is constant.

2. Suppose $f$ is analytic on the set $D$, and suppose $Ref$ is constant on $D$. Is $f$ necessarily constant on $D$? Explain.

3. Suppose $f$ is analytic on the set $D$ and suppose $|f(z)|$ is constant on D. Is $f$ necessarily constant on $D$? Explain.

No hint is found in the text. Please I need a hint and reference to prove the above statements.

-
For 1), consider power series expansion of $f$ (remember - you can differentiate term-by-term). For 2) consider the Cauchy-Riemann equations. For 3) consider the maximum modulus principle. –  William Mar 1 '12 at 10:04
@WNY: the expression of $f$ is not given, what do you think is the power series expansion of $f$? All what we know is that $f$ is any complex valued function. –  Hassan Muhammad Mar 1 '12 at 10:11
It is this: $f(z) = \sum_{n = 0}^\infty a_n(z - z_0)^n$, on some sufficiently small ball around $z_0$ in $D$ and $a_n$ are constant. This is the most fundamental thing that one should know about analytic functions. Now differentiating term-by-term, you get $f'(z) = \sum_{n = 1}^\infty na_n(z - z_0)^{n-1}$. Since $f'(z) = 0$, you must have $na_n = 0$ for all $n\geq 1$. Hence $f(z) = a_0$ - a constant. Perhaps you should review the very basics of power series expansion of complex analytic functions. –  William Mar 1 '12 at 10:45

For 1, if you regard $f$ as a smooth function on $D\subset \mathbb{R}^2$, $f'(z)=0$ implies that the gradient of $f$ is zero, so $f$ must be a constant function.
For 2, since $f=u+iv$ where $u=Re(f)$ and $v=Im(f)$ satisfies Cauchy-Riemann condition, we have $$v_y=u_x=0\mbox{ and } v_x=-u_y=0$$ since $u$ is constant. This implies that $v$ is constant, and as a result $f=u+iv$ is a constant function.
For 3, since $|f|$ is constant, $$\tag{0}|f|^2=u^2+v^2\equiv c$$where $c$ is a constant. If $c=0$, then $f\equiv 0$. So we assume that $c\neq 0$. Differentiate it with respect to $x$ and $y$, we have $$\tag{1} 2(uu_x+vv_x)=0\mbox{ and }2(uu_y+vv_y)=0$$ Again using Cauchy-Riemann condition, $(1)$ can be written as $$\tag{3} uu_x-vu_y=0\mbox{ and }uu_y+vu_x=0.$$ Eliminate $u_y$ in $(3)$, we get $0=(u^2+v^2)u_x=cu_x$ by $(0)$, which implies that $u_x=0$ since $c\neq 0$. Similarly, eliminate $u_x$ in $(3)$, we get $0=(u^2+v^2)u_y=cu_y$ which implies that $u_y=0$. This implies that $u$ is constant. Using part 2, we can conclude that $f$ is a constant function.
What of if $D\subset \mathbb C$ in $1$, how can you go about it? –  Hassan Muhammad Mar 1 '12 at 10:50
What I mean is that you can consider $f$ as a function on $D\subset\mathbb{C}\approx\mathbb{R}^2$. –  Paul Mar 1 '12 at 14:22