Sanbonani guys!!! 

This is MMTC081 (Complex Analysis Module). Here we look into some uncommon problems we have encountered in the course of studying this module. 

 (1) Let $\gamma(t)=[z_1, z_2, z_3]$ be a path where $z_1 = -1$, $z_2 = 1$, $z_3 = i$ and $0 \leq t \leq 1$. Evaluate the path integral.

\[\int_{\gamma^*} \bar{z}\,dz\]


Solution:    

Given the path $\gamma(t)=[z_1,z_2,z_3]$, where $z_1=-1$, $z_2=1$, and $z_3=i$. We note that the possible combination of the points of $\gamma(t)$ are $[z_1,z_2],[z_2,z_3],[z_1,z_3],[z_1,z_2,z_3],[z_1,z_2,z_3,z_1]$ . The path integral of $\bar{z}$ over a path $\gamma$ is given by $\int_\gamma\bar{z}dz$. For a straight-line segment, the parametrization $\gamma(t)=(1-t)z_a+tz_b$, the reveser parametrization $\gamma^*(t)=(1-t)z_b+tz_a$. Our aim is to evaluate the path integrals of these paths beginning with $[z_1,z_2]$. 

(i) Let the segment $\gamma_1$ to be from $z_1=-1$ to $z_2=1$ with its parametrization as $\gamma_1(t)=(1-t)z_1+tz_2\Rightarrow\gamma_1(t)=-1+2t$, where $t\in[0,1]$.  

But our aim is to find the reverse parametrization. We note that the conjugate path $\gamma_1^*$ is simply  $\gamma_1$ traversed backward, this is given as $\gamma_1(t)=(1-t)z_2+tz_1\Rightarrow\gamma_1^*(t)=1-2t$, $t\in[0,1]$. Next, we compute $\int_{\gamma_1^*}$ $\bar{z}dz$. Since $\gamma_1^*$ is $z(t)=1-2t$, $\bar{z}(t)=1-2t$, and $\frac{dz}{dt}=-2$. Thus:  

$\int_{\gamma_1^*}\bar{z},dz=\int_{0}^{1}(1-2t)(-2)dt$, 

$= -2\int_{0}^{1}(1-2t),dt= -2\left[t-t^2\right]_0^1= 0$.

Therefore, the intgeral of $\gamma_1=0$. 

(ii) Next we compute for $[z_2,z_3]$, the reverse parametrization of this is $\gamma_2^*=(1-t)z_3+tz_2=(1-t)i+t=t+i(1-t)$, $dz=(1-i)dt$, and $\bar{z}=\bar{t+i(1-t)}=t-i(1-t)$. Thus 

$\int_{[z_2,z_3]}\bar{z}dz=\int^1_0(t-i(1-t))(1-i)dt=(1-i)\int^1_0(t-i+it)dt$,

$=(1-i)\left[\frac{t^2}{2}-it+i\frac{t^2}{2}\right]^1_0=(1-i)\left[\frac{1}{2}-\frac{i}{2}\right]=\frac{(1-i)^2}{2}=\frac{-2i}{2}=-i$. 

(iii) Now we compute integral over $[z_1,z_3]$, the reverse parametrization is given by 

$\gamma^*_3=(1-t)i-t$, $dz=(-1-i)dt$, and $\bar{z}=\bar{-t+i(1-t)}=-t-i(1-t)$. Thus 

$\int_{[z_1,z_3]}\bar{z}dz=\int^1_0(-t-i(1-t))(-1-i)dt=(-1-i)\int^1_0(-t-i+it)dt$

$=(-1-i)\left[-\frac{t^2}{2}-it+i\frac{t^2}{2}\right]^1_0=(-1-i)(-\frac{1}{2}-i+\frac{i}{2})$

$=(-1-i)\left(-\frac{1}{2}-\frac{i}{2}\right)=\frac{(1+i)(1+i)}{2}=\frac{i^2+2i+1}{2}=\frac{2i}{2}=i.$.

(iv) The integral over $[z_1,z_2,z_3]$ can simply be splitted into two line segments $\int_{[z_3,z_2]}\bar{z}dz+\int_{[z_1,z_2]}\bar{z}dz$. We recall from above computations that 

$\int_{[z_3,z_2]}\bar{z}dz=-i$, and $\int_{[z_1,z_2]}\bar{z}dz=0$. 

Therefore $\int_{[z_1,z_2,z_3]}\bar{z}dz=-i+0=-i$.

(v) Integral over the closed path $[z_1,z_2,z_3,z_1]$ is given by three line-segments

$\int_{[z_1,z_3]}\bar{z}dz+\int_{[z_3,z_2]}\bar{z}dz+\int_{[z_2,z_1]}\bar{z}dz$ and we note from above computations that 

$\int_{[z_1,z_3]}\bar{z}dz=i$, $\int_{[z_3,z_2]}\bar{z}dz=-i$ and $\int_{[z_2,z_1]}\bar{z}dz=0$.

Therefore, $\int_{[z_1,z_2,z_3,z_1]}\bar{z}dz=i-i+0=0$. 

Evaluate Other Paths

$\int_{[z_1,z_2]}\bar{z},dz=0$

$\int_{[z_2,z_3]}\bar{z},dz=-i$

$\int_{[z_1,z_3]}\bar{z},dz=i$

$\int_{[z_1,z_2,z_3]}\bar{z},dz=-i$

$\int_{[z_1,z_2,z_3,z_1]}\bar{z},dz=0$


(2) Find the length of the arch of the curve $\gamma(t)=2t+\frac{2i}{3}t^\frac{3}{2}$, $1\leq{t}\leq 2.$ 

Solution: 

Arc Length of $\gamma(t)$ Given the curve $\gamma(t)=2t+\frac{2i}{3}t^{\frac{3}{2}}$ for $1\leq{t}\leq{2}$, 

we compute its arc length. To do this, we first compute $\gamma'(t)$. Since $\gamma(t)=2t+\frac{2i}{3}t^{\frac{3}{2}}$ then $\gamma'(t)=2+it^{\frac{1}{2}}$.

Then we move to find the magnitude of $\gamma'(t)$. $|\gamma'(t)|=\sqrt{2^2+\left(t^{\frac{1}{2}}\right)^2}=\sqrt{4+t}$.

Now we compute the arc length integral given by,

$L=\int_{1}^{2}\sqrt{4+t}dt.$

Finally we evaluate the integral by let $u=4+t$, so that $du=dt$ and 

$L=\int_{5}^{6}\sqrt{u}$, $du=\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{5}^{6}=\frac{2}{3}\left(6\sqrt{6}-5\sqrt{5}\right).$

$\frac{2}{3}\left(6\sqrt{6}-5\sqrt{5}\right)$.

Which bring us to the end of these problems. Dankie!!!