1
$\begingroup$

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).

enter image description here

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.

  • 1
    Did you try to use $\frac{z}{|z|} = \frac{1}{\overline{z}}$ and integrate $\frac{1}{t - i(1-t)}$ ? – 2017-01-22
  • 0
    @N.H. Thanks. I think I did it. However, the answer terms seems to be very rare. Can you see my edit? – 2017-01-22
  • 0
    $ \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. – 2017-01-22
  • 0
    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. – 2017-01-22
  • 0
    @N.H. Don't worry. I guess I'll try to do the calculation "by hands". Thanks. – 2017-01-22

1 Answers 1

0

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 $$\int \frac{1}{\sqrt{x^2+1}}dx=\log \left(x+\sqrt{x^2+1}\right)+C. $$