Messages

1

MAT334--Lectures & Home Assignments / Re: Residue for FE P3

« on: December 12, 2018, 07:53:50 PM »

I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$


Just asking for a check of this idea.

what i do is since it is a double-pole, i find the coefficient for the degree 2 term of the sine squared function, then i find the coefficient of the degree 1 term for the cosine function, then divide it out.

I don't think you can directly divide them out? Because the coefficient is not that simply decided?

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MAT334--Lectures & Home Assignments / Residue for FE P3

« on: December 12, 2018, 03:27:33 PM »

I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.

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MAT334--Lectures & Home Assignments / Re: FE Sample Question 4 (a)

« on: December 08, 2018, 04:17:00 PM »

Can you approach this method another way without the hint?

I.e) using the three fixed points (choose the last one to be any point that matches the conditions)

Can we choose the third point on the boundary of the domain and image (in this case, both are the unit circle itself), like f(1) = 1?

I don't think you can because in that way you will get a mobius transformation that
$1 \mapsto 1, 0 \mapsto 5, -1 \mapsto -1$, this set of constraints can also be interpreted as the circle passing $1,0,-1$ is mapped to a circle passing $1, 5, -1$, which doesn't satisfy the question?

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MAT334--Lectures & Home Assignments / Sample Final, Q4, (c)

« on: December 06, 2018, 12:15:20 PM »

I just want to ask, if that is the answer. Because the author just took off the absolute value without reason ($z \in \mathbb{C}$ right?) and when dealing with $arg(f'(z))$ I don't understand why the simplification is necessary. Why not just write $arg(f'(z)) = arg(\frac{-24}{(1+5z)^2})$ Why it has to move $\pi$ out?

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MAT334--Lectures & Home Assignments / Re: 3.3 Q7, b,c,d

« on: December 05, 2018, 08:44:34 PM »

Also, is it true that if a mobius transformation maps a circle/a line to a circle/line, then the points in one area in the domain (because there are totally two areas) that is on one side of the circle/line will always maps onto the area in the same side of the circle/line in the codomain?

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MAT334--Lectures & Home Assignments / 3.3 Q7, b,c,d

« on: December 05, 2018, 08:28:32 PM »

I have tried (b) and had some clue about it. Basically I choose the intersection of the original to map to the intersection in the codomain. That means choosing $\{0, \infty\} \mapsto \{0, 1\}$. After that, I can get only a form like $z \mapsto \frac{az}{az+d}$. And then I give it $x \in \mathbb{R}$ and $iy, y \in \mathbb{R}$ to ensure the two condition is satisfied. With that, I can get $ad \in \mathbb{R},\bar{a}d \in \mathbb{R}$. And I arbitrarily chose  a solution and it works.

My question is if there is a general way to do this kind of question? Because in the above, there are too many guessing.

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End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 1

« on: November 30, 2018, 11:52:33 AM »

Ende, what this $n$ soaring in the air means. Probably you forgot parenthesis.

Also $|z|=R$ is definitely overcomplicated since geometric series obviously diverges at each such point

Fixed.
The point is, I cannot see why "geometric series obviously diverges at each such point".

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MAT334--Lectures & Home Assignments / Final Exam Scope?

« on: November 29, 2018, 07:53:08 PM »

What will be covered in the final? Since on the schedule, a conformal mapping is still something marked as "(if permits)".

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End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 1

« on: November 28, 2018, 08:00:40 PM »

This is my solution of part (a).

But it asks for Taylor Series at $0$. You decomposed it at 4.

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End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 1

« on: November 27, 2018, 10:16:47 PM »

Part a).

\begin{align*}
    &f(z) = \frac{1}{(z-(3i+4))(z-(-3i + 4))} = \frac{\frac{1}{8 - 6i}}{(3i+4) - z} - \frac{\frac{1}{8 - 6i}}{(4 - 3i) - z}\\
    &= \frac{\frac{1}{2} \frac{1}{4-3i} \frac{1}{4+3i}}{1 - \frac{z}{4+3i}} - \frac{\frac{1}{2} \frac{1}{4-3i}^2}{1 - \frac{z}{4-3i}}\\
    &= \frac{1}{50} \sum (\frac{z}{4+3i}) ^ n - \frac{1}{2} (\frac{1}{4-3i})^2 \sum (\frac{z}{4-3i})^n
\end{align*}
We need $|\frac{z}{3i+4}| < 1$ and $|\frac{z}{4-3i}| < 1$,
Thus the radius of convergence is $r = 5$, now we consider if the series converges at the $|z | = 5$.

Let $\theta$ be s.t. $\cos \theta = \frac{4}{5}, \sin \theta = \frac{3}{5}$.
Thus, let $t $ be arbitrary,
\begin{align*}
    &f(5e^{it}) = K_1 \sum \frac{5 e^{it}}{5e^{i\theta}} ^ n + K_2 \sum \frac{5 e^{it}}{5e^{-i\theta}} \\
    &= K_1 \sum e^{ni(t-\theta)} + K_2 \sum  e^{ni(t+\theta)}\\
    & = K_1 \sum e^{ni(t-\theta)} + |K_2| \sum  e^{ni(t+\theta)}e^{i2\theta} \\
    & = K_1 \sum e^{ni(t-\theta)} + |K_2| \sum  e^{i(nt + (n+2)\theta)}
\end{align*}
if $f(5e^{it})$ converges, then $Re\{f(5e^{it})\}$ converges, since
\begin{align*}
    Re\{f(5e^{it})\} = K_1 \sum \cos{n(t-\theta)} + |K_2| \sum \cos {(nt + (n+2)\theta)}
\end{align*}
then $\lim_{n \rightarrow \infty}  K_1  \cos{n(t-\theta)} + |K_2|  \cos {(nt + (n+2)\theta)} = 0$, since $K_1 = K_2 $, $\lim_{n \rightarrow \infty}  K_1  \cos{n(t-\theta)} + |K_2|  \cos {(nt + (n+2)\theta)} = \lim_{n \rightarrow \infty} 2K_1 \cos (nt+\theta) \sin (-n\theta - \theta) = 0$, a contradiction.

Thus, on the $|z| = 5$, series does not converge.

Part b).

\begin{align*}
    &f(z) = \frac{1}{(z-(3i+4))(z-(-3i + 4))} = \frac{\frac{1}{8 - 6i}}{(3i+4) - z} - \frac{\frac{1}{8 - 6i}}{(4 - 3i) - z}\\
    &= \frac{1}{z(6i-8)} \frac{1}{1 - \frac{3i+4}{z}} + \frac{1}{z(8-6i)} \frac{1}{1 - \frac{4-3i}{z}} \\
    &= \frac{1}{z(6i-8)} \sum (\frac{3i+4}{z})^n + \frac{1}{z(8-6i)} \sum (\frac{4-3i}{z})^n \\
    &= \frac{1}{(6i-8)} \sum \frac{(3i+4)^n}{z^{n+1}} + \frac{1}{(8-6i)} \sum \frac{(4-3i)^n}{z^{n+1}}
\end{align*}
Still, $|\frac{3i+4}{z}| < 1$ and $|\frac{-3i+4}{z}| < 1$ lead to $|z| > 5$. Thus $R = 5$.

Use same approach as above, we can get
\begin{align*}
    &f(5e^{it}) = K_3 \sum (\frac{5 e^{i\theta}}{5e^{it}}) ^ n \frac{1}{z} + K_4 \sum (\frac{5 e^{i(-\theta)}}{5e^{it}})^n \frac{1}{z} \\
    &= \frac{K_3}{5} \sum e^{ni(\theta - t) - it} + \frac{K_4}{5} \sum  e^{ni(-\theta - t) - it} \\
    &= \frac{K_3}{5} \sum e^{i(n\theta -(n+1)t)} + \frac{K_4}{5} \sum  e^{i(n(-\theta) -(n+1)t)} \\
    &= \frac{K_3}{5} \sum e^{i(n\theta -(n+1)t)} + \frac{K_4}{5} \sum  e^{i(n(-\theta) -(n+1)t)} \\
    &= |\frac{K_3}{5}| \sum e^{i(n\theta -(n+1)t)} e^{i(\theta - \pi)} + |\frac{K_4}{5}| \sum  e^{i(n(-\theta) -(n+1)t)} e^{i\theta} \\
    &= |\frac{K_3}{5}| \sum e^{i((n+1)\theta -(n+1)t - \pi)}  + |\frac{K_4}{5}| \sum  e^{i((n+1)(-\theta) -(n+1)t)} \\
\end{align*}
if $f(5e^{it})$ converges, then $Im\{f(5e^{it})\}$ converges, since
\begin{align*}
    Im\{f(5e^{it})\} = |K_5| \sum -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sum \sin {((n+1)(-\theta) -(n+1)t)}
\end{align*}
then $\lim_{n \rightarrow \infty} |K_5|  -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sin {((n+1)(-\theta) -(n+1)t)} = 0$, since $|K_5| = |K_6|$, $\lim_{n \rightarrow \infty} |K_5|  -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sin {((n+1)(-\theta) -(n+1)t)} = -2|K_5| \lim_{n \rightarrow \infty} \cos (-(n+1) t) \sin ((n+1)\theta) =  0$ a contradiction.

Thus, on the $|z| = 5$, series does not converge.

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MAT334--Lectures & Home Assignments / Re: 3.2 Q1?

« on: November 27, 2018, 01:50:40 PM »

Never mind, I figured out that I can just specifically analyze.
In Q1, I just draw a graph to see that $|z+2|$ can be at most 3 while the numerator is always 1.

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MAT334--Lectures & Home Assignments / 3.2 Q1?

« on: November 27, 2018, 12:05:43 PM »

There is no example in the book.
The only way I know is to translate it into multivariable calculus and use Lagrange multiplier. However, is there an easier way?

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Term Test 2 / Re: TT2 Problem 3

« on: November 25, 2018, 05:30:24 PM »

(Note: At $z = \pi $ or $z = -\pi$, $f(z)$ has pole of order 1, since
$\lim_{z \rightarrow \pi} f(z) = (l'hopital) \lim_{z \rightarrow \pi} \frac{(4z^3-2z\pi^2) \cos^2(z) - \sin(2z)(z^4 - z^2\pi^2)}{sin(2z)} = \infty$, which must be a pole
)

Complete Solution:
Since
$$
f(z) = \frac{z^2(z+\pi)(z-\pi) \cos^2(z)}{\sin^2(z)}
$$
thus,
$f(z)$ has singularities at $z = n\pi$ where $n \in \mathbb{Z}$.
and at each $n\pi$, $\sin^2(z)$ has zero of order 2.

Consider numerator, $z \mapsto z^2(z+\pi)(z-\pi)\cos^2(z)$ has zero of order 2 at $0\pi$, zero of order 1 at $\pi$ and $-\pi$, thus $f(z)$ has a removable singularities at 0, pole of order 1 at $\pi$ and $-\pi$, and pole of order two at $n\pi$ where $|n| \ge 2$.

Consider $g(z) := f(\frac{1}{z})$, since any disc centered at 0 in $g$ will include more than 1 singularities, that means, it is no the case that there exists a small ball, 0 is the only singularity in it. We can conclude $\infty$ is non-isolated singularity for $f$.

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MAT334--Lectures & Home Assignments / Re: 2.6 Q22

« on: November 23, 2018, 07:25:37 PM »

Why "greater than $2 \pi$" does not matter?
I see that you didn't emphasize $\beta$ to be a natural number, which is a bit weird because I don't even know how many singularities there are.

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MAT334--Lectures & Home Assignments / Re: 2.6 Q22

« on: November 22, 2018, 02:12:09 PM »

I don't understand your question.
Isn't it just substituting in the number?