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Quiz 3 / Quiz3 problem 6E
« Last post by Xuefen luo on October 18, 2020, 03:52:58 PM »
Problem: Compute the following line integral:
\begin{align*}\int_{\gamma}^{} |z|^2 \,dz, \end{align*}
where $\gamma$ is the line segment from 2 to 3 + i
\begin{align*}
\end{align*}
Answer:
\begin{align*}
 \gamma(t)&= (1-t)z_0+tz_1\\
 &= 2(1-t)+(3+i)t\\
 &= 2-2t+3t+it\\
 &= 2+t+it \ \ \ \  (0 \leq t \leq 1)\\
 \\
 \gamma'(t)&=1+i\\
 \\
 Let\ f(z)=|z|^2\\
 \\
 f(\gamma(t))&= |\gamma(t)|^2\\
 &=(\sqrt{(2+t)^2+t^2})^2\\
 &=4+4t+t^2+t^2\\
 &=2t^2+4t+4\\\
 \\
 \int_{\gamma}^{} |z|^2 \,dz &= \int_{\gamma}^{} f(z) \,dz\\
 &=\int_{0}^{1} f(\gamma(t))\  \gamma'(t) \,dt\\
 &=\int_{0}^{1} (2t^2+4t+4) (1+i) \,dt\\
  &=(1+i)\int_{0}^{1} (2t^2+4t+4) \,dt \\
   &=(1+i)(\frac{2}{3}t^3 + 2t^2 +4t)|_{0}^{1}\\
    &=(1+i)(\frac{2}{3} + 2 +4)\\
     &=\frac{20}{3}(1+i)
\end{align*}
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Test 1 / Test1-Main-C
« Last post by Qing Lyu on October 16, 2020, 12:29:08 PM »
The attachment files are my questions and answers.
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Test 1 / Re: past test question
« Last post by RunboZhang on October 16, 2020, 09:07:29 AM »
I used ratio test to solve this problem. Since this one has double factorial in its series, so check the limit of $|\frac {a_{n+2}}{a_n}|$ would be helpful. After expanding the limit, you will see most of the terms will be canceled out. What's left is $\frac{1}{n+2}$, and it equals to 0 as n approaches infinity. It also equals to $\frac {1}{R}$. Thus our radius of convergence is infinity.

Another method is to solve this series by splitting it into its even part and odd part. Then use ratio test to get the radius of convergence of both parts. The smaller radius will be the final answer.
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Test 1 / past test question
« Last post by ziyizhou on October 16, 2020, 12:50:59 AM »
how to solve this question step by step?
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Test 1 / Re: TT1 2020S Main Sitting
« Last post by jeyara85 on October 15, 2020, 08:52:42 PM »
Oh okay that makes sense thank you!
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Test 1 / Re: TT1 2020S Main Sitting
« Last post by RunboZhang on October 15, 2020, 06:19:19 PM »
Actually it is $log(2^{\frac {1}{2}})$. But by one property of log function, $log(2^{\frac {1}{2}}) = \frac{1}{2} \cdot log(2)$
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Test 1 / TT1 2020S Main Sitting
« Last post by jeyara85 on October 15, 2020, 05:59:50 PM »
For the step, 6z=1/2log(2)+(3pi/4+2pin)i, why is it log(2) and not log(sqrt2) because I thought it was log(lzl). Also where did the 1/2 come from?
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Test 1 / Re: Test 1 Main E Problem 1
« Last post by Yuyang Wang on October 15, 2020, 03:07:01 PM »
Here is my solution to integral factor in question1(a), it is matter with both x and y so I used another formula:(Nx-My)/(x*M-y*N).
Hopefully this can help you.
And my answer to 1(b) is C=2.
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Test 1 / TT-Main—D
« Last post by Hongye on October 15, 2020, 12:58:38 PM »
Question and answer
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Test 1 / Re: Test 1 Main E Problem 1
« Last post by Domenica16 on October 15, 2020, 09:47:16 AM »
I also had Problem 1 E and I had the same issue, I couldn't seem to solve for the integrating factor. 
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