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### Messages - JUNJING FAN

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1
##### Home Assignment 3 / Re: S2.3 Problem 8
« on: February 05, 2019, 07:57:09 PM »
Find the general solution and then plug to initial conditions
Hello professor, could u give me a hint as to why we have to impose the restriction for absolute value of x to be smaller or equal to 1?

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##### Home Assignment 2 / Re: problem 5 (23)
« on: January 27, 2019, 07:47:30 PM »
There is NO root. You need to parametrize before integration

but, when we parametrize Y in terms of X, don't we have to use $$y^2+x^2=C$$
and thus $$y= +/- \sqrt{C-x^2}$$?

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##### Home Assignment 2 / Need help with S2.2 P1 #4
« on: January 27, 2019, 03:13:52 PM »
I have attached my attempt. However I'm having trouble solving the problem fully with the initial condition provided. Anyone knows how to do it? Or did I make any mistakes along the way?
Thanks.

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##### Home Assignment 2 / Re: problem 5 (23)
« on: January 27, 2019, 12:11:13 PM »
Heller professor
By parametrizing Y in terms of X, do we need to put a plus/minus sign in front of the root? If yes, does that mean when we put down the final solution, we need to include plus and minus as well?

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##### Home Assignment 1 / Re: Home Assignment 1
« on: January 16, 2019, 05:07:33 PM »
Junjing, you are right but I am not sure if anyone but me would be able to read your solution
Okay. Next time I will make it 'skinnier'. Sorry.

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##### Home Assignment 1 / Re: Home Assignment 1
« on: January 16, 2019, 04:36:55 PM »
Not sure if this will work

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##### MAT334--Lectures & Home Assignments / Re: Residue for FE P3
« on: December 12, 2018, 07:02:59 PM »
I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) =$ coefficient of $(z-n\pi)^{1}$ for function $\frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2)$ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2)$
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.
what i do is since it is a double-pole, i find the coefficient for the degree 2 term of the sine squared function, then i find the coefficient of the degree 1 term for the cosine function, then divide it out.

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##### MAT334--Lectures & Home Assignments / Re: Final Exam Scope?
« on: December 12, 2018, 07:00:40 PM »
So we need to know how to do TT1 + TT2 + Q7 ***AND*** the Sample Final correct?

Also, the Sample Final contains questions about the stretch and rotation angle of a mobius transformation. Isn't that part of Chapter 3.4? I don't see anything in 3.3 that discusses rotation and stretch.
same question here
but i guess it never hurts to know how to do it.

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##### MAT334--Lectures & Home Assignments / Re: What's the answer of FE q2 d.e.f?
« on: December 11, 2018, 08:39:41 PM »
Can you explain how you figured it out?
so refer to textbook page 50, the book provides a method. For x just follow the book, for y follow it until the end to prove that y1+y2=0, however the range of y is not limited to be positive in this question. In this case, we will have to expand:

cos(x+iy) = cosx coshy - i sinx sinhy

and realize that sinx is never 0, meaning sinhy will always be there, and sinhy is injective for a positive or negative y of same magnitude.

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##### MAT334--Lectures & Home Assignments / Re: What's the answer of FE q2 d.e.f?
« on: December 11, 2018, 08:28:39 PM »
There are only answers of part a and part b and it seems that we cannot reply to the question anymore.
Thanks
never mind, it is one-to-one. I figured it out.

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##### MAT334--Lectures & Home Assignments / Re: What's the answer of FE q2 d.e.f?
« on: December 11, 2018, 08:17:28 PM »
There are only answers of part a and part b and it seems that we cannot reply to the question anymore.