# Computing $\int_{\partial G}\Im z \thinspace d z$ and $\int_{\partial G}\frac{z}{|z|}dz$ where $G:=\{ z \in \mathbb{D} : \Re z + \Im z > 1 \}.$

I'm doing some complex analysis exercises to not forget what I learned last semester, when I took a first course on the topic. I appreciate if someone could check my work, and help me with one integral I could not do.

Let $\mathbb{D}$ be the unit disk and let $G:=\{ z \in \mathbb{D} : \Re z + \Im z > 1 \}$. Find a convenient parametrization $\gamma$ of $\partial G$ and compute $\int_{\gamma}\Im z \thinspace d z$ as well as $\int_{\gamma}\frac{z}{|z|}dz$.

Okay, so this is $G$. (Sorry for the ugly picture).

The boundary $\partial G$ has then two components: a straight line from $i$ to $1$ and a quarter of the circumference of radius 1 and center 0.

First integral: $\int_{\gamma}\Im z \thinspace d z$. I parametrized the line between $i$ and $1$ as $\gamma_1(t)=t+i(1-t)$, $t \in [0,1]$, and the quarter of circumference as $\gamma_2(t)=e^{it}=\cos t +i\sin t$, $t \in [0,\frac{\pi}{2}]$. Thus, $$\int_{\gamma_1}\Im z \thinspace dz = \int_{0}^{1}(1-t)(1-i)dt = (1-i)(t-\frac{t^2}{2})\Big|_{0}^1=\frac{1}{2}(1-i)$$ and $$\int_{\gamma_2}\Im z \thinspace dz = \int_{0}^{\frac{\pi}{2}}\sin t(-\sin t +i\cos t)dt = \Big(\frac{\sin 2t}{4}-\frac{t}{2}\Big)\Big|_{0}^{\frac{\pi}{2}}+i\Big(\frac{\sin^2 t}{2}\Big)\Big|_{0}^{\frac{\pi}{2}}=i-\frac{\pi}{4}.$$ Hence $$\int_{\gamma}\Im z \thinspace d z = \int_{\gamma_1}\Im z \thinspace dz + \int_{\gamma_2}\Im z \thinspace dz = \frac{1}{2}(1+i) -\frac{\pi}{4}.$$

Second integral: $\int_{\gamma}\frac{z}{|z|}dz$. Okay, for the quarter of circumference I got no problem. I let $\gamma_2(t)=e^{it}$, $t \in [0,\frac{\pi}{2}]$ and obtain $$\int_{\gamma_2}\Im z \thinspace dz = \int_{0}^{\frac{\pi}{2}}\frac{e^{it}}{|e^{it}|}ie^{it}dt = i\int_{0}^{\frac{\pi}{2}}e^{2it}dt=\frac{i}{2i}e^{2it} \Big|_{0}^{\frac{\pi}{2}}=\frac{1}{2}(e^{i\pi}-1).$$ However, if I let $\gamma_2(t)=t+i(1-t)$, $t \in [0,1]$, then \begin{equation*} \begin{aligned} \int_{\gamma_1}\frac{z}{|z|}dz = (1-i)\int_{0}^{1}\frac{t+i(1-t)}{|t+i(1-t)|}dt &= (1-i)\int_{0}^{1}\frac{|t+i(1-t)|(t+i(1-t))}{|t+i(1-t)|^2}dt \\ &= (1-i)\int_{0}^{1}\frac{|t+i(1-t)|(t+i(1-t))}{t^2+(1-t)^2}dt \end{aligned} \end{equation*} Then I don't know how to proceed. I think there might be an easier approach. Maybe finding a convenient parametrization, as the exercise suggest.

Questions:

1) Can anyone see if what I did is correct? If not, where is the mistake?

2) How can I compute the last integral?

I appreciate any hints, comments or suggestions.

• Did you try to use $\frac{z}{|z|} = \frac{1}{\overline{z}}$ and integrate $\frac{1}{t - i(1-t)}$ ? – user171326 Jan 22 '17 at 0:41
• @N.H. Thanks. I think I did it. However, the answer terms seems to be very rare. Can you see my edit? – positrón0802 Jan 22 '17 at 2:17
• $\frac{z}{|z|}=\frac{1}{\bar{z}}$ is not correct. Since $|z|^2=z\bar{z}$, $\frac{z}{|z|}=\frac{|z|}{\bar{z}}$ is correct. But, perhaps it is no use. – ts375_zk26 Jan 22 '17 at 5:58
• Right, I'm so stupid >< sorry @positrón0802, my hint was useless. But I don't think there is an easy way of computing your integral, since the function your are trying to integrate is not holomorphic, so essentially trying to do calcul "by hands" like you did seems to be the best solution. – user171326 Jan 22 '17 at 11:41
• @N.H. Don't worry. I guess I'll try to do the calculation "by hands". Thanks. – positrón0802 Jan 22 '17 at 18:43

Perhaps there are no convenient parametrizations. Using parametrization $\gamma_1(t)=t+i(1-t), t\in [0,1],$ we calculate it. $$\int_{\gamma_1}\frac{z}{|z|}dz = (1-i)\int_{0}^{1}\frac{t+i(1-t)}{|t+i(1-t)|}dt= (1-i)\int_{0}^{1}\frac{t+i(1-t)}{\sqrt{t^2+(1-t)^2}}dt,$$ \begin{align} \int_{0}^{1}\frac{t}{\sqrt{t^2+(1-t)^2}}dt&=\frac{\sqrt{2}}{4}\int_{-1}^{1}\frac{s+1}{\sqrt{s^2+1}}ds\quad (s=2t-1)\\ &=\frac{\sqrt{2}}{2}\int_{0}^{1}\frac{1}{\sqrt{s^2+1}}ds=\frac{\sqrt{2}}{2}\log (1+\sqrt{2}). \end{align} Similarly $$\int_{0}^{1}\frac{1-t}{\sqrt{t^2+(1-t)^2}}dt=\frac{\sqrt{2}}{2}\log (1+\sqrt{2}).$$ Recall $$\int \frac{1}{\sqrt{x^2+1}}dx=\log \left(x+\sqrt{x^2+1}\right)+C.$$