$f(z)$ analytic, then $|f(z)|$ or $\arg(z)$ constant, then $f(z)$ constant

Question

Ir order to show that $f(z)$ is constant I need to show that its partial derivatives are all $0$, that is:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 0$$

I tried this:

If $|f(z)|$ is constant, then $\sqrt{u^2+v^2} = c \implies$

$$\frac{2u}{2\sqrt{u^2+v^2}}=0\implies u = 0$$ and $$\frac{-2v}{2\sqrt{u^2+v^2}}=0 \implies v = 0$$

That's strange because I didn't even use Cauchy-Riemann and I tought it would be necessary. Also, how to do the part of the argument?

Answer 1

A possible problem with your argument is that $u$ being related to $v$.

In a simplified example with $u=x$, $v=y$, let $\sqrt{x^2+y^2}=1$. When we take a "derivative" of this with respect to $x$, we consider $y$ as a dependent variable, not an independent one. Of course, this does not yield $x=y=0$.

So, in this one, we need Cauchy-Riemann.

Answer 2

Careful! You're treating the functions $u,v$ as variables. You should be differentiating the modulus $|f|$ with respect to $x$ and $y$, not with respect to $u$ and $v$. So, e.g., $$ 0 = \frac{\partial |f|}{\partial x}(x,y) = \frac{uu_x + vv_x}{|f(x,y)|} $$ and so long as $|f|\neq 0$ (why may you assume this?) you may conclude that the vector $(u,v)$ is orthogonal to the vector $(u_x,v_x)$. Likewise, $(u,v)$ is orthogonal to $(u_y,v_y)$. Now you are in a position to use Cauchy-Riemann ...

For the argument, recall from polar coordinates that $\tan \arg(w) = y/x$. Then differentiate as above and apply Cauchy-Riemann.

Answer 3

The equation $\sqrt{u^2+v^2} = c$ implies the pair $(u,v)$ is constrained to lie on the circle of radius $c$ centered at $0$. It is impossible for $u$ to change while $v$ remains fixed and at the same time stay on that circle. But the derivative of $\sqrt{u^2+v^2}$ with respect to $u$, which you computed, is not valid unless you're talking about $u$ changing while $v$ is fixed.

Now imagine a tiny circle of (positive) radius $\varepsilon\ll1$ centered at $0$, and $dz$ as a tiny complex number with $|dz|=\varepsilon$, and let $dz$ change as $\arg(dz)$ goes from $0$ to $2\pi$. What then happens to $f(z_0+dz)$, where $z_0$ is in the interior of the domain of $f$? If $f'(z_0)\ne 0$, then $f'(z)$ is close to $f'(z_0)$ whenever $z$ is close to $z_0$, so $f(z_0+dz)-f(z_0)\approx f'(z_0)\,dz$ goes around an approximate circle of radius $|f'(z_0)|\cdot\varepsilon$ centered at $f(z_0)$. It cannot go around that tiny circle while remaining on the big circle $\sqrt{u^2+v^2} = c$.

Heuristics like that are worth knowing about, but are not what is considered logically rigorous. For the logically rigorous version, Cauchy–Riemann will serve.

Answer 4

Here's another proof: Suppose $|f(z)|$ is constant, then so is $|f(z)|^2$. Since constant functions are analytic, then we have that \begin{align} 0=\frac{\partial}{\partial \bar z}|f(z)|^2= \frac{\partial f}{\partial \bar z} \bar f + f\frac{\partial \bar f}{\partial \bar z} = \frac{\partial f}{\partial \bar z}\bar f + f\overline{\frac{\partial f}{\partial z}}. \end{align} Since $f$ is analytic, then $\partial_{\bar z} f =0$ which means \begin{align} f\overline{\frac{\partial f}{\partial z}} =0 \ \ \Rightarrow \ \ \frac{\partial f}{\partial z} =0. \end{align} Hence $f$ is constant.

Answer 5

Use the Cauchy-Riemann equations in polar coordinates! If you're not familiar, see this question: Proof of Cauchy Riemann Equations in Polar Coordinates

From here, basically the standard argument applies. If one is constant, the other is also constant, meaning the function is constant.

Answer 6

Suppose $f'(a) \neq 0$ for some $a$ in $\Bbb{C}$. Then as $\delta \to 0$ $f(a + \delta) = f(a) + \delta f'(a) + o(\delta)$. Since $|f(a + \delta) - (f(a) + \delta f'(a))|$ is $o(\delta)$, and $|f(a) - (f(a) + \delta f'(a))| = |\delta f'(a)|$ is not $o(\delta)$, for small $\delta$ $f(a + \delta)$ is closer to $f(a) + \delta f'(a)$ than $f(a)$ is. Let $\delta = r \alpha$, where $\alpha f'(a)$ is an outward pointing normal vector to the circle $|z| = |f(a)|$ at $f(a)$ and $r$ is a positive real. As $r \to 0$, $f(a + \delta)$ is closer to $f(a) + \delta f'(a)$ than $f(a)$ is. But $f(a)$ is clearly the closest point to $f(a) + \delta f'(a)$ on the circle $|z| = |f(a)|$, so this contradicts the assumption of constant modulus.

I like this proof because it doesn't really use any machinery of complex analysis-- it's more explicitly geometric.