### Recent Posts

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##### Quiz 7 / Q7: Wednesday Sitting
« Last post by Ziyi Wang on March 22, 2020, 02:51:12 PM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the  given annulus: $z^3-3z+1$ in ${1<|z|<2}$
Answer: $f(z)=z^3-3z+1$
On the circle $|z|=1$
$$|z^3+1| \leq |z^3|+|1|=2$$
$$|-3z| = |3z| = 3$$
So $|z^3+1|<|-3z|$. By Roche's Theorem, $f(z)$ and $g(z)=-3z$ have the same number of zeroes with $|z|=1$.
Since $g(z)$ has one zero with $|z|=1$, we know that $f(z)$ has one zero with $|z|=1$.
On the circle $|z|=2$
$$|z^3|=8$$
$$|-3z+1| \leq |3z|+|1| = 7$$
So $|-3z+1|<|z^3|$. By Roche's Theorem, $f(z)$ and $g(z)=z^3$ have the same number of zeroes with $|z|=2$.
Since $g(z)$ has three zeroes with $|z|=1$, we know that $f(z)$ has three zeroes with $|z|=1$.
$$3-1=2$$
Therefore, $f(z)=z^3-3z+1$ has two zeroes in ${1<|z|<2}$.
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##### Chapter 6 / Re: Laplace Equation Section 6.4
« Last post by Victor Ivrii on March 21, 2020, 09:55:28 AM »
It means that $\Theta$ is $2\pi$-periodic.
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##### Chapter 6 / Re: Laplace Equation Section 6.4
« Last post by Kuangyi Tang on March 19, 2020, 07:45:52 PM »
I understand now, there is another constraint $$\Theta'(0)=\Theta'(2\pi)$$, but what does this condition mean physically?
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##### Chapter 6 / Laplace Equation Section 6.4
« Last post by Kuangyi Tang on March 19, 2020, 07:22:02 PM »
In section 6.4, the topic about separation of variables.
I don't exactly understand why the constraint  $$\Theta(0)=\Theta(2\pi)$$ gives the Eigen functions $\Theta_{s,n}(\theta)=sin(n\pi \theta)$and $\Theta_{c,n}(\theta)=cos(n\pi \theta)$. Shouldn't the condition produce the Eigen functions $\Theta_{s,n}(\theta)=sin(\frac{n \pi \theta}{2})$and $\Theta_{c,n}(\theta)=cos(\frac{n\pi \theta}{2})$ instead?
Since previously a similar problem $$\Theta'' + \lambda \Theta = 0$$ $$\Theta(0)=\Theta(\alpha)$$ give the Eigen functions $sin(\frac{n\pi \theta}{\alpha} )$ $cos(\frac{n\pi \theta}{\alpha} )$ and the Eigenvalue$\lambda_n = (\frac{n \pi}{\alpha})$

P.S. I couldn't get the Quercus discussion to work
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##### Test 2 / Topics covered in TT2; schedule of the 4 sittings; and Final arrangement
« Last post by zeyang zhang on March 18, 2020, 10:20:15 PM »
1.The old coverage was:
It will be 20 pts worth, containing 4 problems, each 5 pts worth, based on Chapters 4--6 of the online Textbook. and a 2.5pt bonus

2.What is the schedule of the 4 sittings in TT2?

3.BTW, what will be the format of Final exam? Will it be the same with Quiz and TT2? We should have the info by today......
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##### Chapter 3 / Re: 3.1 question 13 typo
« Last post by Victor Ivrii on March 16, 2020, 10:08:38 AM »
Thanks. Fixed
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##### Chapter 3 / 3.1 question 13 typo
« Last post by Yan Zhou on March 15, 2020, 11:49:24 PM »
The range is $\frac{1}{2} < |z| < \frac{3}{2}$. Not $\frac{1}{2} < |z| < 1$.
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##### Chapter 3 / 3.1 question 8
« Last post by Yan Zhou on March 15, 2020, 11:28:52 PM »
$$2z^4 - 2iz^3 + z^2 + 2iz -1$$
find roots for upper half line.
http://forum.math.toronto.edu/index.php?topic=1591.0
here is the link for previous quiz solution.
But I do not get why argf(x) changes -2$\pi$
I tried to analyze the change sign for both Im(f) and Re(f), This is what I got:
x<-1, Im(f) > 0
-1<x<0, Im(f) <0
0<x<1, Im(f) >0
x>1,Im(f) <0
for real part:
x<-$\frac{1}{2}$ ,Re(f) > 0
-$\frac{1}{2}$< x < $\frac{1}{2}$, Re(f) < 0
x > $\frac{1}{2}$, Re(f) > 0
Then first, f moves from first quadrant to third quadrant  through second quadrant, then f moves back to first quadrant through second quadrant and f ends up in fourth quadrant. Therefore, I think arg(f) changes at most $-\pi$.

Can anyone help figure out which part is wrong?
Thank you!
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##### Chapter 3 / 3.1 question 5
« Last post by Yan Zhou on March 15, 2020, 10:17:20 PM »
$$f(z) = z^9 + 5z^2 + 3$$
I have difficulty in figuring out how f moves on $iy$.
The following is my steps.
$$f(iy) = iy^9 - 5y^2 + 3$$
y moves from R to 0. Im(f) > 0, f always lie in first or second quadrant.
f(0) = 3 on the real axis.
$$f(iR) = iR^9 - 5R^2 + 3$$
$$arg(f) = arctan(\frac{R^9}{-5R^2 + 3})$$
As R goes to $\infty$, arg(f) goes to $\frac{\pi}{2}$,
Then arg(f) changes from $\frac{\pi}{2}$ to 0, then $\Delta arg(f) = -\frac{\pi}{2}$
I am not sure about arctan part, should it goes to $\frac{\pi}{2}$ or  $-\frac{\pi}{2}$
and if there is any other mistakes, thanks for pointing out!
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##### Chapter 5 / Re: Section 5.1, 5.2 - ok to consider delta distribution?
« Last post by Victor Ivrii on March 10, 2020, 08:35:56 AM »
Then you need to explain more or less rigorously, what is $\delta$-distribution and how Fourier transform is defined for it. Yes, online textbook covers this topic and much more than your physics class but in more advanced chapters, we do not cover
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