On the subject of holomorphic functions on an open disc, D.

Question

The question I am pondering over is an interesting one:

If $f(z) = u + iv$ is holomorphic on an open disc $D$, and the range of $f$ lies in either a straight line or a circle, prove that $f$ is constant on $D$.

Do I just need to just Cauchy-Riemann?

Answer 1

Case I. The values of $f=u+iv$ lie on a straight line.

Thus there exists $a,b, c\in\mathbb R$, with $(a,b)\ne (0,0)$, such that $$ au+bv=c. $$ Thus, if we set $$g(z)= (a-ib)f(z)=(a-ib)(u+iv)=(au+bv)+i(av-bu)=c+i(av-bu),$$ then $g$ is also holomorphic and $$ \mathrm{Re}\,g(z)=c. $$ But is the real part of $g=\alpha+i\beta$ is constant, then due to Cauchy-Riemann equations $$ 0=\alpha_x=\beta_y, \quad 0=\alpha_y=-\beta_x, $$ the imaginary part of $g$ is also constant, and hence $g$ is constant.

Case II. The values of $f$ lie on a circle.

Say $$ f(z)\in \{w: |w-w_0|=r\}. $$ Set $$ g(z)=f(z)-w_0. $$ Then the values of $g=u+iv$ line on the unit circle, and hence $\lvert g \rvert=r$, which implies that $g$ is constant, and so is $f$.

Answer 2

It's a basic fact in complex analysis that a non constant holomorphic function is open...

Answer 3

You can take the Cauchy-Riemann equations and the partial derivatives of either $u^2+v^2=r^2$ or $au+bv=c$ to form a homogenous 4x4 system for the partial derivatives of $u$ and $v$. Then show that it is regular ($r>0$ resp. $a^2+b^2\ne0$ required, $r=0$ is trivial) to conclude that the functions are constant.

Answer 4

Here is an elementary method using only cauchy riemann equations:

For the first part, rotate to assume the function has image contained in the reals. Then by cauchy riemann equations the derivative is zero since the partial derivatives of the imaginary part are obviously zero.

For the second part translate to assume that $|f|$ is constant. So use $c=|f|^2=f \bar{f}$ to get that either $f$ is zero or $\bar{f}$ is holomorphic. Then apply cauchy riemann equations to prove that if $f$ and $\bar{f}$ are both holomorphic then the derivative of $f$ is zero.

Of course this trivially follows from the open mapping theorem also.