Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – user12918723509187 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. – user12918723509187 Mar 1 '12 at 10:45
up vote 4 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.