MAT334-2018F > Term Test 1

TT1 Problem 4 (noon)

(1/1)

Victor Ivrii:
Calculate directly integral
$\displaystyle{\int_L \frac{dz}{z}}$ where $L$ is  the counter-clockwise oriented boundary of the square $\{(x,y)\colon -1<x<1, -1<y<1\}$:

oighea:
Since the square is a piecewise simple closed curve, and that the integrand is of the form $\frac{1}{z-p}$, here $p = 0, p \in \mathrm{In}\,L$, the integral should be $i2\pi$.

We will verify by calculating the integral directly without Green's Theorem. Define $L$ as four separate smooth curves:

$\gamma_1 = (1-i)+it, \gamma'_1 = i, t \in [0,2]$
$\gamma_2 = (1+i)-t, \gamma'_2 = -1, t \in [0,2]$
$\gamma_3 = (-1+i)-it, \gamma'_3 = -i, t \in [0,2]$
$\gamma_4 = (-1-i)+t, \gamma'_4 = 1, t \in [0,2]$

The integrands will be:

$\displaystyle A_1 = \int^2_0 \frac{i}{(1-i)+it}dt = i[\frac{1}{i}i\ln(1-i+it)]^2_0 = [\ln(1-i+it)]^2_0 = \ln(1+i) - \ln(1-i) = \ln\left(\frac{1+i}{1-i}\right) = \ln\left(\frac{e^{i\pi/4}}{e^{-i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_2 = \int^2_0 \frac{-1}{(1+i)-t}dt = -[-1\ln(1+i-t)]^2_0 = [\ln(1+i-t)]^2_0 = \ln(-1+i) - \ln(1+i) = \ln\left(\frac{-1+i}{1+i}\right) = \ln\left(\frac{e^{3i\pi/4}}{e^{i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_3 = \int^2_0 \frac{-i}{(-1+i)-it}dt = -i[\frac{1}{-i}\ln(-1+i-it)]^2_0 = [\ln(-1+i-it)]^2_0 = \ln(-1-i) - \ln(-1+i) = \ln\left(\frac{-1-i}{-1+i}\right) = \ln\left(\frac{e^{-3i\pi/4}}{e^{3i\pi/4}}\right) = \ln{e^\frac{3i\pi}{2}} = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_4 = \int^2_0 \frac{1}{(-1-i)+t}dt = [\ln(-1-i+t)]^2_0 = [\ln(-1-i+t)]^2_0 = \ln(1-i) = \ln(-1-i) = \ln\left(\frac{1-i}{-1-i}\right) = \ln\left(\frac{e^{-i\pi/4}}{e^{-3i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$

Summing the integrals of the curves $A$ together, we obtain

$A = i\pi(\frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2})=i2\pi$.