# Integral with differential is purely imaginary

Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $\gamma$, then $\int_\gamma\overline{f(z)}f'(z)dz$ is purely imaginary.

I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $\int_a^b\overline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?

• Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it? – achille hui Oct 5 '13 at 19:27
• @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is. – Paul S. Oct 5 '13 at 19:30
• $2^{nd}$ Hint: what is $\frac{d}{dt} |f(z(t))|^2$? – achille hui Oct 5 '13 at 19:34
• @achillehui It is $\overline{f(z(t))}f'(z(t))z'(t)+f(z(t))\overline{f'(z(t))}\overline{z'(t)}$. This can be written as $2\Re[\overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part). – Paul S. Oct 5 '13 at 19:44
• Well okay, so the integral of the real part is $\dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes! – Paul S. Oct 5 '13 at 19:50

Let be $$f = u + iv$$, $$D =$$ the interior of $$\gamma$$. Using Cauchy-Riemann and Green: $$\int_\gamma\bar{f(z)}f'(z)\,dz = \int_\gamma(u - iv)(u_x + iv_x)\,(dx + idy)$$ $$= \int_\gamma\left[(uu_x + vv_x)+(uv_x - vu_x)i\right](dx + idy)$$ $$= \int_\gamma(uu_x + vv_x)\,dx - (uv_x - vu_x)\,dy + i\int_\gamma\cdots$$ $$= \int_\gamma(uu_x + vv_x)\,dx + (uu_y + vv_y)\,dy + i\int_\gamma\cdots$$ $$= \frac12\iint_\gamma\partial_x(u^2 + v^2)\,dx + \partial_y(u^2 + v^2)\,dy + i\int_\gamma\cdots$$ $$= \frac12\iint_{D}\left[\partial_x\partial_y(u^2 + v^2) - \partial_y\partial_x(u^2 + v^2)\right]\,dxdy + i\int_\gamma\cdots$$ $$= 0 + i\int_\gamma\cdots$$
Let $r(z)=|f(z)|$ and $\theta(z)=\arg(f(z))$. Start with simplifying.
$$\int_\gamma\overline{f(z)}f'(z)\ dz=\int_\gamma|f(z)|^2\frac{f'(z)}{f(z)}\ dz=\int_\gamma e^{2\ln|f(z)|}\frac{f'(z)}{f(z)}\ dz=\int_\gamma e^{-2i\arg(f(z))}e^{2\ln f(z)}\frac{f'(z)}{f(z)}\ dz=\int_\gamma e^{-2i\arg(f(z))}f(z)f'(z)\ dz=\int_\gamma e^{-2i\theta(z)}r(z)e^{i\theta(z)}[r'(z)e^{i\theta(z)}+ir(z)\theta'(z)e^{i\theta(z)}]\ dz=\int_\gamma r(z)r'(z)\ dz+i\int_\gamma r^2(z)\theta'(z)\ dz=\frac{1}{2}r^2(z)\Big|_\gamma+i\int_\gamma r^2(z)\ d[\theta(z)]$$
The term for the real part is clearly 0 as $\gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $\theta(z)$ are real. Thus the original integral's value must be purely imaginary.