# Suppose that $f$ is analytic on a close curve γ. Prove or disprove $\int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.

Suppose that $f$ is analytic on a close curve γ. Prove or disprove

$$\int_\gamma \overline{f(z)}f'(z)dz$$

is purely imagine.

I know that $f$ is analytic on a close curve, then

$$\int_\gamma f(z) dz=0$$

I tried an example with $\gamma =e^{it}$ with $0\leq t\leq 2\pi$, I often get the real part of the integral equal zero.

I tried let $f=u+iv$ so $f'=u_x+iv_x$. Since $f=u+iv$, $\overline{f}=u-iv$

$$\overline{f(z)}f'(z)=uu_x+vv_x+i(uv_x-vu_x)$$

But this doesn't get me anywhere. Any help would be greatly appreciated.

$$0 = \int_{\gamma} d\, |f|^2 = \int_{\gamma} \left(\overline{f} \cdot f' dz + f\cdot \overline{f'} d\overline{z}\right) = 2 \textrm{Re} \int_{\gamma}\overline{f} \cdot f' dz.$$

• How did you get the last step? Commented Jun 7, 2021 at 21:25
• @ExponentBOT From $w + \overline w = 2 \mathrm{Re}(w)$ for $w \in \mathbb C$.
– WimC
Commented Jun 8, 2021 at 4:26

If $f = u + iv$, then

\begin{align}\text{Re}(\overline{f(z)}f'(z)\, dz) &= \text{Re}([(uu_x + vv_x) + i(uv_x - vu_x)](dx + i\, dy))\\ & = (uu_x + vv_x)\, dx + (-uv_x + vu_x)\, dy . \end{align}

Since $u_x = v_y$ and $u_y = -v_x$,

$$(uu_x + vv_x)\, dx + (-uv_x + vu_x)\, dy = (uu_x + vv_x)\, dx + (uu_y + vv_y)\, dy = d\frac{u^2 + v^2}{2}.$$

Thus $\text{Re}(\overline{f(z)}f'(z)\, dz)$ is an exact differential, which implies

$$\int_{\gamma} \text{Re}(\overline{f(z)}f'(z)\, dz) = 0.$$

Therefore, $\int_\gamma \overline{f(z)}f'(z)\, dz$ is purely imaginary.

• 1. How do you know that $Re(\overline{f(z)}f;(z)\,dz)$ is exact? Commented Jan 24, 2018 at 20:56
• 2 How does (1) imply the integral is 0? Commented Jan 24, 2018 at 20:57
• @UsernameUnknown (1) because it's equal to $d\phi$ where $\phi = \frac{u^2 + v^2}{2}$, and (2) by the fundamental theorem of line integrals.
– kobe
Commented Jan 24, 2018 at 21:00