### 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.

### Topics - Weiyin Wu

Pages: 
1
##### Quiz 4 / TUT0401
« on: February 16, 2020, 08:26:05 PM »
Find the radius of convergence of the given power series.
$$\sum_{k=1}^{\infty} k(z-1)^{k}$$
$$\frac{1}{R}=\lim_{k\to\infty} |\frac{a_{k+1}}{a_{k}}|=\lim_{k\to\infty} |\frac{k+1}{k}|=1$$
$$R=1$$

2
##### Quiz 3 / TUT0401 QUIZ3
« on: February 06, 2020, 01:57:50 PM »
Let $\gamma_{1}$ be the semicircle from 1 to -1 through i and $\gamma_{2}$ the semicircle from 1 to -1 through -i
Compute $\int_{\gamma_{1}} z^2 dz$ and  $\int_{\gamma_{2}} z^2 dz$
Can you account for the fact that they are equal?
$$r_{1}(t) = e^{it} (0\leq t \leq\pi)$$
$$r'_{1}(t) = ie^{it}$$
$$\int_{\gamma_{1}} z^2 dz = \int_{0}^{\pi} (e^{it})^{2}\cdot ie^{it}dt$$
$$=i\int_{0}^{\pi} e^{3it}dt$$
$$=\frac{1}{3}(e^{3i\pi}-e^{0})$$
$$=\frac{1}{3}(\cos (3\pi) +i\sin (3\pi)-1)$$
$$=\frac{-2}{3}$$
Similarly,
$$r_{2}(t) = e^{it} (-\pi\leq t \leq 0)$$
$$r'_{2}(t) = ie^{it}$$
$$\int_{\gamma_{2}} z^2 dz = \int_{0}^{-\pi} (e^{it})^{2}\cdot ie^{it}dt$$
$$=\frac{1}{3}(e^{-3i\pi}-e^{0})$$
$$=\frac{1}{3}(\cos (-3\pi) +i\sin (-3\pi)-1)$$
$$=\frac{-2}{3}$$
Yes, $\int_{\gamma_{1}} z^2 dz = \int_{\gamma_{2}} z^2 dz$ since $\gamma_{1}$ and $\gamma_{2}$ are in the same direction, $z^{2}$ is analytic and the region (a circle) is close.

3
##### Quiz 2 / TUT0401 QUIZ2
« on: January 30, 2020, 09:07:15 PM »
Find the value of the given expression:
$$(1+i)^i$$
$$= e^{ln(1+i)^i}$$
$$= e^{i\cdot ln(1+i)}$$
$$= e^{i\cdot (ln|1+i|+i\cdot arg(1+i))}$$
$$= e^{i\cdot (ln\sqrt {2}+i(\frac{\pi}{4}+2k\pi))} (k\in \mathbb{Z})$$
$$= e^{i\cdot (ln\sqrt {2})} \cdot e^{-(\frac{\pi}{4}+2k\pi))}$$
$$=(\cos{ln\sqrt {2}}+i\sin{ln\sqrt {2}})\cdot e^{-(\frac{\pi}{4}+2k\pi))}$$

4
##### Quiz 2 / TUT0401
« on: January 30, 2020, 08:28:50 PM »
Find the limit of each function at the given point, or explain why it does not exist.
$f(z) = |1 - z|^2$ at $z_0 = i$
$$\lim_{z\to z_0} f(z) = \lim_{z\to z_0} |1 - z|^2 = |1 - i|^2 = 1^2 + (-1)^2 = 2$$

Pages: