Show that $z^n+c$ is harmonic, $c\in\mathbb C$.

Question

I would like to know how to show $$f(z) = z^n+c$$ for $c\in \mathbb C$ is harmonic over $D =\{|z|\leq r\}$.

I know that if I express $z = x+iy$, then I can have $f=u+iv$, where $u$ and $v$ will be polynomials. We know polynomials are are analytic over $D$ and hence $f$ is analytic. Since $f$ is analytic, $u$ and $v$ are harmonic.

Is my reasoning correct? Is there a good/faster way to show this rigorously?

Answer 1

Your reasoning is basically assuming the answer. Your answer amounts to "$f$ is a polynomial, so it is holomorphic, so its components are harmonic". The answer is correct, but it doesn't have any work in.

I would probably feel obliged to prove that $z^n + c$ is holomorphic directly. To do this, note that the sum and product of holomorphic functions is holomorphic, so it suffices to show that $z$ and $c$ are holomorphic functions. $c$ is obviously: it differentiates to $0$. $z$ is obviously: it differentiates to $1$. Tiny amount of work to do there: $$\lim_{h \to 0} \frac{x+h - x}{h} = \lim_{h \to 0} 1 = 1$$

Answer 2

It can be proved by the Laplacian in polar coordinates, i.e. $$ u_{rr}+\frac{u_r}{r}+\frac{u_{\theta\theta}}{r^2}=0\tag1 $$ Since $$ f(z)=r^ne^{in\theta}+c=r^n\cos{n\theta}+ir^n\sin{n\theta}+c=u+iv $$ There is $$ u_{rr}=n(n-1)r^{n-2}\cos{n\theta} $$ And $$ \frac{u_{r}}{r}=nr^{n-2}\cos{n\theta} $$ $$ \frac{u_{\theta\theta}}{r^2}=-n^2r^{n-2}\cos{n\theta} $$ Hence $u$ satisfies $(1)$ and is harmonic. So is $v$.