$f(z)$ is analytic in a region that contains closed curve $\gamma$ . Show that
$\int \limits_{\gamma}{\overline{f(z)}f'(z)}\, \mathrm{d}z$
is purely imaginary.
We have to show that $Re[\int \limits_{\gamma}{\overline{f(z)}f'(z)}\, \mathrm{d}z] = 0$
how to show that?
On
Suppose $\gamma$ is continuously differentiable and parametrized over, say, $[0,1]$. Then, by definition, your integral is: $\int_0^1 \overline f(\gamma(t)) f'(\gamma(t)) \gamma'(t) dt$ which, in turn, by the chain rule, equals: $\int_0^1 \overline (f\circ \gamma)(t) (f\circ \gamma)'(t) dt.$ Putting $g(t) = f\circ \gamma(t)$, we finally get: $\int_0^1 \overline g(t) g'(t) dt$
Now, if we write $g(t) = u(t) + iv(t)$, the integral can be rewritten as: $\int_0^1 (u - iv)(u' + iv')dt$ (where I omitted the variable t for clarity). The real part of this integral is: $\int_0^1 (uu' + vv')dt = \frac12 \int0^1 (u^2 + v^2)'dt $ $= u^2(1) + v^2(1) - u^2(0) - v^2(0) = 0, $ since $\gamma(1) = \gamma(0)$ (as $\gamma$ is closed)
On
Writing $f = u + iv$, $z = x + iy$ and using C-R equations and the Green's theorem: $$\eqalign{\int_\gamma\overline{f(z)}f'(z)\,dz &= \int_\gamma(u-iv)(u_x+iv_x)\,(dx+idy) = \int_\gamma\left[(uu_x+vv_x)+(uv_x-vu_x)i\right](dx+idy)\cr &= \int_\gamma(uu_x+vv_x)\,dx-(uv_x-vu_x)\,dy + i\int_\gamma\cdots\cr &= \int_\gamma(uu_x+vv_x)\,dx+(uu_y+vv_y)\,dy + i\int_\gamma\cdots\cr &= \frac12\iint_\gamma\partial_x(u^2+v^2)\,dx + \partial_y(u^2+v^2)\,dy + i\int_\gamma\cdots\cr &= \frac12\iint_{\mathrm{int}(\gamma)}\left[\partial_x\partial_y(u^2+v^2) - \partial_y\partial_x(u^2+v^2)\right]\,dxdy + i\int_\gamma\cdots\cr &= 0 + i\int_\gamma\cdots }$$
Let $f=u+iv$. Then $$\overline{f(z)}f'(z)dz=(u-iv)(u_x+iv_x)(dx+i dy)\ .$$ The real part of this amounts to $$\omega:=(u u_x+vv_x)dx+(vu_x-uv_x)dy=(u u_x+vv_x)dx+(vv_y+uu_y)dy\ ,$$ using the CR equations. It follows that $$\omega={1\over2}d(u^2+v^2)\ ,$$ hence $\int_\gamma\omega=0$.