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

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?

11

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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