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### Topics - Yiheng Bian

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##### 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
$$\int_{\Upsilon}e^z dz = \int_{0}^{1}f(r(t))r'(t)dt = \int_{0}^{1}e^{tz_{0}}z_{0} dt = z_{0}(\frac{1}{z_{0}}e^{z_{0}}- \frac{1}{z_{0}})= e^{z_{0}} - 1$$

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##### 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
$$\text{limit is zero}$$

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##### 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
$$k=0, \theta=0,z=1(\cos0+i\sin0)=1$$
$$k=1, \theta=\frac{\pi}{4},z=1(\cos(\frac{\pi}{4})+I\sin{\frac{\pi}{4}})$$
$$k=2, \theta=\frac{\pi}{2},z=1(\cos(\frac{\pi}{2})+i\sin{\frac{\pi}{2}})$$
$$k=3, \theta=\frac{3\pi}{4},z=1(\cos(\frac{3\pi}{4})+i\sin{\frac{3\pi}{4}})$$
$$k=4, \theta=\pi,z=1(\cos\pi+i\sin\pi)$$
$$k=5, \theta=\frac{5\pi}{4},z=1(\cos(\frac{5\pi}{4})+i\sin{\frac{5\pi}{4}})$$
$$k=6, \theta=\frac{3\pi}{2},z=1(\cos(\frac{3\pi}{2})+i\sin{\frac{3\pi}{2}})$$
$$k=7, \theta=\frac{7\pi}{4},z=1(\cos(\frac{7\pi}{4})+i\sin{\frac{7\pi}{4}})$$

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

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