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Topics - Yiheng Bian

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1
Quiz 3 / TUT5101 QUIZ3
« on: February 06, 2020, 12:34:14 AM »
$$
\int_{\Upsilon}e^z dz
$$
$$
\text{line from 0 to } z_{0}$$
Therefore
$$
r(t)= tz_{0}
$$
$$
r'(t) = z_{0} (0\leq t \leq 1)
$$
And since
$$
f(z)=e^z
$$
$$
f(r(t))= e^{z_{0}}
$$
So
$$
\int_{\Upsilon}e^z dz =  \int_{0}^{1}f(r(t))r'(t)dt = \int_{0}^{1}e^{tz_{0}}z_{0} dt = z_{0}(\frac{1}{z_{0}}e^{z_{0}}- \frac{1}{z_{0}})= e^{z_{0}} - 1
$$

2
Quiz 2 / TUT5101 QUIZ2
« on: January 30, 2020, 12:42:13 AM »
Question:
$$
h(z)=\frac{Log(z)}{z}
$$
We know when
$$
 z \rightarrow \infty
$$
$$
Log(z) \rightarrow \infty
$$
So we can use L'Hopital rules and get
$$
\frac{1}{z}
$$
So
$$
\text{limit is zero}
$$


3
Quiz 1 / TUT5101
« on: January 26, 2020, 01:14:20 AM »
Find all solutions of the given equation:
$$
z^8 =1
$$
We know
$$
z=r(\cos\theta +I\sin\theta)
$$
$$
z^8=r^8(\cos8\theta +I\sin8\theta)
$$
Then
$$
1=1+0i=1(\cos2k\pi + I\sin2k\pi)
$$
So we have
$$
r^8(\cos8\theta +I\sin8\theta) = 1(\cos2k\pi + I\sin2k\pi)
$$
Therefore we get
$$
r=1
$$
$$
8\theta=2k\pi
$$
$$
\theta= \frac{k\pi}{4}
$$
So when
$$
k=0, \theta=0,z=1(\cos0+i\sin0)=1
$$
$$
k=1, \theta=\frac{\pi}{4},z=1(\cos(\frac{\pi}{4})+I\sin{\frac{\pi}{4}})
$$
$$
k=2, \theta=\frac{\pi}{2},z=1(\cos(\frac{\pi}{2})+i\sin{\frac{\pi}{2}})
$$
$$
k=3, \theta=\frac{3\pi}{4},z=1(\cos(\frac{3\pi}{4})+i\sin{\frac{3\pi}{4}})
$$
$$
k=4, \theta=\pi,z=1(\cos\pi+i\sin\pi)
$$
$$
k=5, \theta=\frac{5\pi}{4},z=1(\cos(\frac{5\pi}{4})+i\sin{\frac{5\pi}{4}})
$$
$$
k=6, \theta=\frac{3\pi}{2},z=1(\cos(\frac{3\pi}{2})+i\sin{\frac{3\pi}{2}})
$$
$$
k=7, \theta=\frac{7\pi}{4},z=1(\cos(\frac{7\pi}{4})+i\sin{\frac{7\pi}{4}})
$$

4
Quiz-4 / TUT0101 QUIZ4
« on: October 18, 2019, 08:41:21 PM »
$$
\text{The question is } 9y''+9y'-4y=0
$$
$$
9r^2+9r-4=0
$$
$$
\text{So r1}=\frac{1}{3}
$$
$$
r2=\frac{-4}{3}
$$
$$
\text{So we can get y}=c_1e^{\frac{t}{3}}+c_2e^{\frac{-4t}{3}}
$$

5
Quiz-3 / TUT0101 QUIZ3
« on: October 12, 2019, 12:57:19 AM »
Answer

6
Quiz-2 / TUT0101
« on: October 05, 2019, 11:39:56 AM »
ANSWER

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