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?
 A: Let be $f = u + iv$, $D =$ the interior of $\gamma$. Using Cauchy-Riemann and Green:
$$
\int_\gamma\bar{f(z)}f'(z)\,dz
= \int_\gamma(u - iv)(u_x + iv_x)\,(dx + idy)
$$
$$
= \int_\gamma\left[(uu_x + vv_x)+(uv_x - vu_x)i\right](dx + idy)
$$
$$
= \int_\gamma(uu_x + vv_x)\,dx - (uv_x - vu_x)\,dy + i\int_\gamma\cdots
$$
$$
= \int_\gamma(uu_x + vv_x)\,dx + (uu_y + vv_y)\,dy + i\int_\gamma\cdots
$$
$$
= \frac12\iint_\gamma\partial_x(u^2 + v^2)\,dx + \partial_y(u^2 + v^2)\,dy + i\int_\gamma\cdots
$$
$$
= \frac12\iint_{D}\left[\partial_x\partial_y(u^2 + v^2) - \partial_y\partial_x(u^2 + v^2)\right]\,dxdy + i\int_\gamma\cdots
$$
$$
= 0 + i\int_\gamma\cdots
$$
A: Let us write $f = u + iv$. Then from analyticity of $f$ we have:
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and }\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}  $$
Now,
$$\int_{\gamma} \bar{f}f'dz = (u-iv)(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}) (dx + idy)\\
=\int_{\gamma}[(u\frac{\partial u}{\partial x} + v\frac{\partial v}{\partial x}) + i(u\frac{\partial v}{\partial x} - v\frac{\partial u}{\partial x})][dx + idy]$$
After multiplication we find it's real part to be:
$$\int_{\gamma} (u\frac{\partial u}{\partial x}dx -u\frac{\partial v}{\partial x}dy) + (v\frac{\partial v}{\partial x}dx + v\frac{\partial u}{\partial x}dy)$$
After replacing $ - \frac{\partial v}{\partial x}$ to $\frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial x}$ to $\frac{\partial v}{\partial y}$ we get:
$$\int_{\gamma} (u\frac{\partial u}{\partial x}dx + u\frac{\partial u}{\partial y}dy) + (v\frac{\partial v}{\partial x}dx + v\frac{\partial v}{\partial y}dy)$$
The integrand is easily seen to be an exact differential of $\frac{1}{2} (u^2 + v^2)$ and hence the integral over any closed curve $\gamma$ is $0$.
A: How about this solution
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.
