Solutions To Problems in Complex Analysis[MAT306 Test]

Solution to MAT306[Complex Analysis II] Test

Below is a solution to questions of MAT306 complex analysis II test.
The topics treated include contour integration, connected regions and singular points.
Also treated are solutions to problems on  line integral, cauchy integral formula, complex laurent expansion series(Theorem and example) and lastly how to determine the number of roots in a circle of a complex equations.
Below are the questions and solutions.


Note:Please kindly use the comment box or contact form on the homepage of this blog for any comment, corrections and observations, because i cannot guarantee a 100% accuracy due to typographic errors and others errors made during the process of solving.

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Questions

(1a) Define the following terms

(i) contour

(ii) connected regions

(iii) singular points 

(1b) Evaluate the integral $\int^{(2,4)}_{(2,3)}(2y+x^2)dx+(3x-y)dy$ along the parabola $x=2t$ and $y=t^2+3$.

(1c). Evaluate $\int\frac{2z}{(z^2-16)(z-i)}dz$ if $C$ is the circle $\{|z|=3\}$
(2a) State the laurent series expansion theorem
(b) Compute $\int\frac{1+2z^2}{z^3+z^5}dz$ for any contour $C$ lying entirely in

$\{0<|z|<1\}$ and surrounding $0$.



Solutions

(i) contour: This is a path in the complex plane over which contour integration is

 performed to compute a contour integral.

(ii) A connected region is said to be simply connected if any simple close-curve can be

 shrunk to a point continuously in the set. If the region is connected but not simply, then

 it is multiply connected.

(iii) Singular points of an analytic function is a point at which an analytic function

$f(z)$ is not analytic. i.e at which  $f'(z)$ fails to exists is called a singular point or

 singularity of the function.

(1b) Since parabola is given at $x=2t$ then $dx=2dt$ and $y=t^2+3$, $dy=2tdt$, we

 substitute the parabolic values and their respective derivatives into the integral.

$\int_{(2,3)}^{(2,4)}(2(t^2+3)+(2t)^2)2dt+(3(2t)-(t^2+3))2tdt$

$\int_{(2,3)}^{(2,4)}(2t^2+6+4t^2)2dt+(6t-t^2-3)2tdt$

Since the points are $(2,4),(2,3)$ and the integral is homogeneous we simply pick the

 $(2,2)$ as our limits.

$\int^2_2(12t^2+12)dt+(12t^2-2t^3-6t)dt$

$\int^2_2(24t^2-2t^3-6t+12)dt=\frac{24t^3}{3}-\frac{2t^4}{4}-\frac{6t^2}{2}+12t|^2_2$

$8t^3-\frac{t^4}{2}-3t^2+12t|^2_2=[8(2)^3-\frac{2^4}{2}-3(2)^2+12(2)]-[8(2)^3-\frac{2^4}{2}-3(2)^2+12(2)]=0$

and that is the required line integral.

(1c) Using the cauchy integral formula The function $\frac{2z}{(z-4)(z+4)(z-i)}$ is not

 analytic at $z=4,-4,i$, Now $C$ is the circle given by $|z|=3$, the point $z=0$ lies

 within $C$.

$\therefore$ $\int_c\frac{2z}{(z-4)(z+4)(z-i)}dz=\int\frac{2z}{\frac{(z-4)(z+4)}{(z-i)}}$

=$\frac{2z}{(z-4)(z+4)}|_{z=i}$

$\frac{2i}{(i-4)(i+4)}=\frac{2i}{i^2+4i-4i-16}=\frac{2i}{-1-16}$ 

 By cauchy integral formula

$\frac{-2i}{17}.2\pi{i}=\frac{4\pi}{17}$

(2a) The laurent series expansion theorem states that if a function is analytic on two

 concentric circles $C_1$ and $C_2$ centred at $z_0$ and in the interior region between

 them, then there is an infinite series expansion with positive and negative powers of

 $z-z_0$ about $z=z_0$ representing this function in this region is  called the laurent series.

 This is written as

$f(z)=\sum^\infty_{n=0}a_n(z-z_0)^n+\sum^\infty_{n=1}\frac{b_n}{(z-z_0)^n}$ where $a_n=\frac{1}{2\pi{i}}\oint_{c_2}\frac{f(\xi)}{(z-z_0)^{n+1}}d\xi$ $n=0,1,2,...$
and

$b_n=\frac{1}{2\pi{i}}\oint_{c_1}f(\xi)(\xi-z_0)^{n+1}d\xi$ $n=1,2,3,...$


(2b)   $\int\frac{1+2z^2}{z^3+z^5}dz=\int_c\frac{1+2z^2}{z^3(1+z^2)}$ where

 $z^3=0\Rightarrow{z=0}$ and $1+z^2=0\Rightarrow{z^2=-1}\Rightarrow{z=i}$

hence the roots is $z=0,i$ and the pole lies inside the circle at $z=0$ hence by the cauchy

 integral formula

$\int\frac{1+2z^2}{\frac{1+z^2}{z}}=\int\frac{f(z)}{z-a}dz=2\pi{f(a)}$

$=2\pi{i}[\frac{1+2z^2}{1+z^2}]_{z=0}=2\pi{i}[1]=2\pi{i}$ which is the required solution to

 the integral.