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### Messages - Yuechen Huang

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##### Quiz-7 / Re: Q7 TUT 5101
« on: November 30, 2018, 04:12:50 PM »
$1. f(z)=2z^{4}-2iz^{5}+z^{2}+2iz-1$
$f(x)=2x^{4}-3ix^{3}+x^{2}+2ix-1$
$x: [-\infty, \infty]$
$f(-\infty)\rightarrow \infty$
$f(\infty)\rightarrow \infty$ arg$(f(z))=0$

2. $Re^{it}$   $0\leq t \leq \pi$
$f(t)=2R^{4}e^{i4t}- 2iR^{3}e^{i3t}+R^{2}e^{i2t}+2iRe^{it}-1$
$0\leq 4t \leq 4\pi$   arg$(f(z))=4\pi$

The net change of argument is $4\pi$ , so that there are four solutions

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##### Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 03:54:47 PM »

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##### Quiz-5 / Re: Q5 TUT 0101
« on: November 02, 2018, 04:59:54 PM »
z² (1 − cos(z))=0

z = 0 or cos (z) = 1

z = 0 or 𝑧 = 2𝑘𝜋
When z = 0

let 𝑓(𝑧) = z², ℎ(𝑧) = 1− cos (𝑧)

then 𝑓(0) = 0, 𝑓′(0) = 2𝑧 = 0, 𝑓″(0) = 2 ≠ 0

thus order = 2

ℎ(0)=0, ℎ′(0)= 2 sin(0)=0, h″(0)= cos(𝑧)≠0

thus order = 2

Therefore, order (0) = 4

When 𝑧 = 2𝑘𝜋 (𝑘≠0)

let 𝑓(𝑧) = z² and ℎ(𝑧) = 1−cos(𝑧)

𝑓(𝑧) = z² = (2𝑘𝜋)² ≠ 0

thus order = 0

ℎ(𝑧)= 1− cos(𝑧)=0, ℎ′(𝑧)= sin(𝑧)=0, ℎ″= cos(𝑧)=0

thus order = 2
Therefore order (2𝑘𝜋) = 2, 𝑘 ≠ 0

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##### Quiz-5 / Re: Q5 TUT 0102
« on: November 02, 2018, 04:36:44 PM »

f(z) = e^z - 3e^z - 4 = 0

Let $w=e^z$, then

w^2 - 3w - 4 = 0 \\
(w-4)(w+1) = 0 \Rightarrow w = 4 \space or \space w = 1 \\
e^z = 4 \space or \space e^z = -1 \\
z = \log4 \space or \space z = \log(-1) \\

When $e^z = 4$, the order is 1

f'(z) =2e^{2z} - 3e^{z} = 2 \times 4^2 - 3 \times 4 \neq 0

When $e^z = -1$, the order is 1

f'(z) =2e^{2z} - 3e^{z} = 2 \times (-1)^2 - 3 \times (-1) \neq 0

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##### Quiz-4 / Re: Q4 TUT 0202
« on: October 26, 2018, 06:53:54 PM »

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##### Quiz-4 / Re: Q4 TUT 0301
« on: October 26, 2018, 06:18:30 PM »
First we have

\int\limits_{\mid z+1 \mid = 2}\frac{z^2}{4-z^2}\,dz = \int\limits_{\mid z+1 \mid = 2}\frac{z^2}{(2+z)(2-z)}\,dz

Point $z=2$ is outside of the circle $\mid z+1 \mid = 2$ and Point $z=-2$ is inside of the circle $\mid z+1 \mid = 2$

\int\limits_{\mid z+1 \mid = 2}\frac{z^2/(z-2)}{(z+2)}\,dz = \int\limits_{\mid z+1 \mid = 2}\frac{z^2/(z-2)}{z-(-2)}\,dz

This gives us function $f(z)$

f(z) = \frac{z^2}{2-z} \Rightarrow f(-2) = \frac{(-2)^2}{2-(-2)} = \frac{4}{4} = 1

So Cauchy's Formula gives us

\int\limits_{\mid z+1 \mid = 2}\frac{z^2}{4-z^2}\,dz = 2\pi if(-2) = 2\pi i

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