### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - oighea

Pages: [1] 2
1
##### Quiz-6 / Re: Q6 TUT 0203
« on: November 18, 2018, 02:17:49 AM »
This proof does not require the direct evaluation of Laurent/Power series, but instead relies on algebraic and calculus manipulation.

Let $f(z)$ be a function analytic on the punctured disk of radius $R$ centered on $z_0$: $\{z | 0 < |z - z_0| < R\}$, and has a pole of order $l$ @ $z_0$.

Then, $\displaystyle f(z) = \frac{H(z)}{(z-z_0)^l}$, where $H$ is analytic and nonzero on all of the disk $\{z | 0 \leq |z - z_0| < R\}$.

Therefore, we express $\displaystyle H(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + ... = \sum_{k=0}a_k(z-z_0)^k$

We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(z-z_0)^l - l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = \frac{H'(z)(z-z_0)^l}{(z-z_0)^{2l}} - \frac{l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = l\frac{H(z)}{(z-z_0)^{2l - l + 1}} - \frac{H'(z)}{(z-z_0)^{2l - l}}$
We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}$

It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(z-z_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)} - l\frac{1}{(z-z_0)^1}$

From now on, we can see that the residue could be $-l$, but considering $\frac{H'(z)}{1H(z)}$, we conclude that the $H$ being analytic and nonzero on the entire disk implies $H'$ being analytic as well, and finally $\frac{H'(z)}{H(z)}$, and so that fraction has no principal part. This leaves the maximum negative degree of the principal part to be 1.

We conclude that the residue has to be $-l$.

2
##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P3
« on: November 04, 2018, 05:43:36 AM »
As $z^3 \tan(\pi z)\cot^2(\pi z^2)$ involves quotients of trigonometric functions, we obtain:

$\displaystyle f(z) = z^3 \frac{\sin(\pi z)}{\cos(\pi z)}\frac{\cos^2(\pi z^2)}{\sin^2(\pi z^2)}$.

Requirements for being a singular point
• $\cos (\pi z)$ is zero, which follows $\tan (\pi z)$ is a simple pole at that point. $\cos (\theta)$ is zero where $\theta$ is a half-integer multiple of $\pi$.
• $\sin (\pi z^2)$ is zero, which follows $\cot (\pi z^2)$ is a pole of order 2 at that point. $\sin (\theta)$ is zero where $\theta$ is an integer multiple of $\pi$.
Singular points at $\mathbb{C}$
This function is singular at all points such $\cos(\pi z)=0$ and all points such $\sin(\pi z^2)=0$.
• Case 1: $z$ is a half-integer. Then $z = k + \frac{1}{2}, k \in \mathbb{Z}$. Then $\pi z$ will be a half-integer multiple of $\pi$. Then $\tan(\pi z)$ will have a simple pole at that point since $\sin(\pi z) \neq 0, \cos(\pi z) = 0$ (Simple pole)
• Case 2: $z^2$ is an integer, $k$. Then $z = \sqrt{k}, k \in \mathbb{Z}$, and $z$ is either on the real or imaginary axis. Then $\pi z^2$ will be an integer multiple of $\pi$, so $\cot^2(\pi z^2)$ will have a pole up to order 2 at that point since $\sin(\pi z^2) = 0$ at that denominator.
• Case 2a: $z^2$ is a negative integer, $-k$ where $k \in \mathbb{N}$. Then $z = i\sqrt{k}$, on the imaginary axis. Only $\sin^2(\pi z^2)$ is zero. (Double pole)
• Case 2b: $z^2$ is a positive integer, $k$, but $z$ is irrational.  Only $\sin^2(\pi z^2)$ is zero. (Double pole)

Pages: [1] 2