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### Messages - Yan Zhou

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1
##### Chapter 3 / 3.1 question 13 typo
« on: March 15, 2020, 11:49:24 PM »
The range is $\frac{1}{2} < |z| < \frac{3}{2}$. Not $\frac{1}{2} < |z| < 1$.

2
##### Chapter 3 / 3.1 question 8
« 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!

3
##### Chapter 3 / 3.1 question 5
« 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!

4
##### Chapter 4 / 4.1 typos
« on: March 04, 2020, 02:45:36 PM »
The solutions to equation (11) should be $k_{1,2} = \pm \frac{\pi n}{l}i$. There is a missing sign of $\pm$.

In the second term of equation (13), $B_{n}sin(\frac{c\pi nt}{l})$ has an extra letter A.

5
##### Test 1 / past test 2019 day problem 3B
« on: February 21, 2020, 12:18:28 PM »
I think there are some typo in the answer of u(x,t).
$$u(x,t) = 6sin(x+3t) + 6sin(x-3t)$$ as $x>3t$,
$$u(x,t) = 6sin(x+3t) + 6sin(2(x-3t))$$ as $t<x<3t$.

6
##### Test 1 / past test 2016F problem 2
« on: February 15, 2020, 10:28:52 PM »
I believe that in question (a), the equation underlined should be $$-\frac{1}{2}[e^{-\frac{1}{2}(x+t-\tau)^2} + e^{-\frac{1}{2}(x-t+\tau)^2}]$$
The rest is correct.

7
##### Chapter 2 / Re: 2.2 home assignment question 18
« on: February 12, 2020, 12:36:51 PM »
Yes, I just find out that it is different from textbook, and I know how to do it now.
By the way, there is a typo in question 16 which should be $$\sum_{n=1}^{\infty} n(z-1)^{n-1}$$ instead of $$\sum_{n=1}^{\infty} (z-1)^{n-1}$$

In section 2.3,
question 5 should be $$\int_{0}^{2\pi} \frac{d\theta}{2+cos\theta}$$ instead of $1+cos\theta$
question 8 should be $$\int_{0}^{\pi}\frac{d\theta}{1+(sin\theta)^2}$$ the range is from 0 to $\pi$ instead of $2\pi$
question 9 should be "joining $1-i$ to $1+i$".

8
##### Chapter 2 / Re: 2.3 question 3
« on: February 10, 2020, 08:11:40 PM »
I found some differences between textbook and home assignment, some of them are typos, and some do not affect the questions but the answers. Should we always follow the questions on the textbook?

9
##### Chapter 2 / Re: 2.3 question 3
« on: February 10, 2020, 07:46:22 PM »
I see. The question on the home assignment is a little bit different from the textbook.

10
##### Chapter 2 / 2.3 question 3
« on: February 10, 2020, 07:22:49 PM »
$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z}$$
Since -2 lies in $|z+1| = 2$, Cauchy theorem gives that
$$\frac{1}{2\pi i} \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{z-(-2)} = \frac{-2}{2-(-2)} = -\frac{1}{2}$$
then $$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = -\pi i$$
However, I checked the answer of textbook and it says the answer is $2\pi i$, I am confused about where i did wrong.

11
##### Chapter 2 / 2.2 home assignment question 18
« on: February 10, 2020, 04:42:13 PM »
Find the closed form for  the given power series.

$$\sum_{n=2}^{\infty}n(n-1)z^{n}$$

hint: divide by $z^{2}$

I tried the hint but still have no idea.

12
##### Chapter 2 / 2.3 textbook
« on: February 09, 2020, 05:52:15 PM »
I am quite confused about the value of $2+sin\theta$. We know $sin\theta = \frac{1}{2i}(z - \frac{1}{z})$,
it says that $2 + sin\theta = 2 +(\frac{1}{2}i)(z-\frac{1}{z})$. Is that a typo?

13
##### Chapter 1 / 1.5 question 23
« on: February 02, 2020, 03:06:29 PM »
23. Show that $F(z) = e^{z}$ maps the strip $S = \{ x+iy: -\infty < x < \infty, -\frac{\pi}{2} < y < \frac{\pi}{2}\}$ onto the region $D = \{w = s+it: s \geqslant 0, w \neq 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary  of $D$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Imz = \frac{\pi}{2}\}$ and $\{Imz = -\frac{\pi}{2}\}$

To prove onto, I want to show, $\forall w \in D, \exists z \in S$, such that $F(z) = e^{z} = e^{x + iy} = w = s +it$.
So I need to find $x,y$ such that $s = e^{x}cosy, t = e^{x}siny$. I got answear $x = \frac{ln(s^{2} + t^{2})}{2}, y = arcos\frac{s}{\sqrt{s^{2} + t^{2}}}$But I don't know if they are correct. Can anyone help with this question and the following questions as well?

14
##### Chapter 2 / S2.4 online textbook
« on: January 20, 2020, 06:33:20 PM »
Does anyone know why is there a -1/4 in the middle equation in picture 1?
From section 2.3, we have the equation on the second picture.
So I think it is a typo. If it is not, please let me know! Thanks in advance.
PS: Does anyone know how to use latex in the post?

15
##### Chapter 2 / Re: S2.2P Problem 2 (6)
« on: January 20, 2020, 06:16:26 PM »
I have attached my solution to equation (6). I hope it may be helpful. I am confused with the following IVP problem as well. Maybe someone else can help.

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