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**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 $\frac{1}{2} < |z| < 1$.

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The range is $\frac{1}{2} < |z| < \frac{3}{2}$. Not $\frac{1}{2} < |z| < 1$.

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$$2z^4 - 2iz^3 + z^2 + 2iz -1$$

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!

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

$$f(z) = z^9 + 5z^2 + 3$$

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!

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!

5

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 $t<x<3t$.

$$u(x,t) = 6sin(x+3t) + 6sin(x-3t) $$ as $x>3t$,

$$u(x,t) = 6sin(x+3t) + 6sin(2(x-3t))$$ as $t<x<3t$.

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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.

The rest is correct.

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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 $1+i$".

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 $1+i$".

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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?

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I see. The question on the home assignment is a little bit different from the textbook.

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$$\int_{|z+1| = 2} \frac{zdz}{4-z^{2}} = \int_{|z+1| = 2} \frac{\frac{zdz}{2-z}}{2+z} $$

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.

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.

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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.

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

hint: divide by $z^{2}$

I tried the hint but still have no idea.

Thanks in advance.

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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?

it says that $2 + sin\theta = 2 +(\frac{1}{2}i)(z-\frac{1}{z})$. Is that a typo?

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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?

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?

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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?

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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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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