Evaluate the integrals $\int_\gamma Im\zeta d\zeta$ and $\int_\gamma\dfrac{\zeta}{|\zeta|}d\zeta$. Let $G := \{ z \in \mathbb{D}: Rez+Imz>1 \}$ and let $\gamma=\partial G$. Evaluate the integrals $\int_\gamma Im\zeta d\zeta$ and $\int_\gamma\dfrac{\zeta}{|\zeta|}d\zeta$.
I am not really shore how to do this.
I know:


*

*I have to define the region of intregation, which is $\partial G$, this is the unitare sphere with a line from i to 1.

*I have to define he paths. One would be the segmento from i to 1 and the other would be part of the sphere

*then if I define well the paths I can integrate in each path and the sum the results.
But I am having trouble with defining the paths. In the first one $f := Im\zeta$.
Thanks for the help.
 A: The path $\gamma$ has two parts $\gamma_1$ and $\gamma_2$, which can be parameterized as follows. I presume the path is meant to be oriented counterclockwise (you didn’t say), but this will only affect the sign of the result.
On $\gamma_1$, put $\zeta(t)=t+(1-t)i$ for $0\leq t\leq 1$. Then $d\zeta =(1-i)\; dt$. The integrand should be obvious in each case.
On $\gamma_2$, put $\zeta(t)=e^{it}$ for $0\leq t\leq \pi/2$. Then $d\zeta = ie^{it}\; dt$, and again the integrand should be clear in each case.
Then you can write $\int_{\gamma} = \int_{\gamma_1} + \int_{\gamma_2}$.
A: The first one is much easier than the other one by using the following theorem from complex analysis:
$$\int_{\partial D} f \:dz = 2i \iint_D \frac{\partial f}{\partial \bar{z}} \: dx\wedge dy$$
Using $\operatorname{Im}(z) = \frac{z-\bar{z}}{2i}$ we have that
$$\int_\gamma \operatorname{Im}(z) \:dz = 2i \iint_{G} -\frac{1}{2i} \:dx\wedge dy = -\left(\frac{\pi}{4}-\frac{1}{2}\right) = \frac{2-\pi}{4}$$
For the second one we can use the same theorem but we will have to evaluate the partial derivative from definition:
$$\frac{\partial}{\partial \bar{z}}\left(\frac{z}{|z|}\right) = \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right)\left(\frac{x}{\sqrt{x^2+y^2}}+ \frac{iy}{\sqrt{x^2+y^2}}\right)$$
$$ = \frac{1}{2}\left(\frac{1}{\sqrt{x^2+y^2}}-\frac{x^2}{(x^2+y^2)^{\
\frac{3}{2}}} - \frac{ixy}{(x^2+y^2)^{\
\frac{3}{2}}} + \frac{ixy}{(x^2+y^2)^{\
\frac{3}{2}}} - \frac{1}{\sqrt{x^2+y^2}} + \frac{y^2}{(x^2+y^2)^{\
\frac{3}{2}}}\right)$$
$$= \frac{1}{2}\frac{y^2-x^2}{(x^2+y^2)^{\frac{3}{2}}}$$
leaving us with the integral
$$\int_\gamma \frac{z}{|z|}\:dz = i\iint_G \frac{y^2-x^2}{(x^2+y^2)^{\frac{3}{2}}}\:dx\wedge dy$$
Now notice that the integrand is negative under the interchange $y \leftrightarrow x$, and the region of integration $G$ is symmetric across the line $y=x$, therefore we can conclude that the integral is $0$ by the properties of odd functions.
$$\int_\gamma \frac{z}{|z|}\:dz = i\iint_G \frac{y^2-x^2}{(x^2+y^2)^{\frac{3}{2}}}\:dx\wedge dy = 0$$
