Take the 2-minute tour ×
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.

I want to show $f$ is entire given the following conditions: $f:\mathbb{C}\to\mathbb{C}$, $f$ is real differentiable, and $f$ is conformal when $f'$ is not $0$.

share|improve this question

1 Answer 1

Real differentiability at ${\bf z}_0\in{\mathbb R}^2$ means that one has $$f({\bf z}_0+{\bf Z})-f({\bf z}_0)=L.{\bf Z}+o(|{\bf Z}|)\qquad({\bf Z}\to{\bf 0}),$$ where $L$ is a certain linear map. In the case at hand this $L$ necessarily has the form $$\bigl[L\bigr]=\lambda\left[\matrix{\cos\phi & -\sin\phi \cr \sin\phi& \cos\phi \cr}\right]=:\left[\matrix{a& -b \cr b& a\cr}\right]\ ,\qquad \lambda\geq0, \quad a,b\in{\mathbb R}\ .$$ Changing from vectorial notation to complex notation we therefore have $$f(z_0+Z)-f(z_0)=(aX-bY)+i(bX+aY)+o(|Z|)=(a+ib) Z+o(|Z|)\qquad (Z\to0)\ ,$$ and this implies that $f$ has a complex derivative at $z_0$ given by $f'(z_0)=a+ib$ .

Since this holds for all $z_0\in{\mathbb C}$ the given function $f$ is indeed entire, and we didn't even need to assume that $f$ is $C^1$.

share|improve this answer

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.