Prove integral of an even complex function over any circle with center zero is zero.

Question

I need to prove integral of an even complex function over any circle is zero.

I know if a function is even, then its antiderivative is odd (if no constant terms), and if we change the orientation, then the integral changes sign for vector field. But how can I prove this?

Any help will be appreciated.

Answer 1

Let $f(z)$ be a continuous complex function on $\mathbb{C}$, such that $f(-z)=f(z) \ \forall z \in \mathbb{C}$

Then $$\int_{|z|=R}f(z)dz=0$$

Where $|z|=R$ is the circle of radius $R>0$ parameterized by $z=Re^{2\pi it}$ for $0\leq t\leq1$.

For simplicity sake, denote such circle by $\gamma$.

There are two different ways (that I can think of) of going about this. Ill walk you through a more elementary and tedious method, and leave the second as an exercise for you. (Which is actually much simpler).

We want to exploit the symmetry of $\gamma$.

Note, $$\int_{\gamma}f(z)dz=\int_0^1f(\gamma(t))\gamma'(t)dt$$

$$=\int_0^{1/4}f(\gamma(t))\gamma'(t)dt+\int_{1/4}^{1/2}f(\gamma(t))\gamma'(t)dt+\int_{1/2}^{3/4}f(\gamma(t))\gamma'(t)dt+\int_{3/4}^1f(\gamma(t))\gamma'(t)dt$$

Each term above is evaluated on a different quadrant of $\gamma$. (I.e. the first contour integral is over the first quadrant of $\gamma$, the second integral over the second quadrant, etc).

Since $\gamma$ is a circle with center $(0,0)\in \mathbb{C}$ and radius $R>0$, it's clear that

$$\gamma([0,\frac {1}{4}])=-\gamma([\frac {1}{2},\frac {3}{4}])$$

Since $f(-z)=f(z) \ \forall z\in \mathbb{C}$, it follows that,

$$f(\gamma([0, \frac {1}{4}]))=f(-\gamma([\frac {1}{2},\frac {3}{4}]))=f(\gamma([\frac {1}{2}, \frac {3}{4}]))$$

Also, note $\frac {d\gamma(t)}{dt}=2\pi iRe^{2\pi i t}$, and thus,

$$\gamma'([0,\frac{1}{4}])=-\gamma'([\frac{1}{2},\frac {3}{4}])$$

Finally we see,

$$f(\gamma([0,\frac {1}{4}]))\gamma'([0,\frac {1}{4}])=-f(\gamma([\frac {1}{2}, \frac {3}{4}]))\gamma'([\frac {1}{2}, \frac {3}{4}])$$

In which it follows that,

$$\int_0^{1/4}f(\gamma(t))\gamma'(t)dt=-\int_{1/2}^{3/4}f(\gamma(t))\gamma'(t)dt\implies\int_0^{1/4}f(\gamma(t))\gamma'(t)dt+\int_{1/2}^{3/4}f(\gamma(t))\gamma'(t)dt=0$$

The same argument holds for the intervals $[\frac {1}{4}, \frac {1}{2}]$ and $[\frac {3}{4},1]$, in which the claim follows.

The above method exploits too many symmetries, thus requiring more work than needed. Instead we can consider the change of variables $t\mapsto t+\frac {1}{2}$. Try this yourself!