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### Topics - 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 / 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.

8
##### 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.

Thanks in advance.

9
##### 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?

10
##### 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?

11
##### 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?

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