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Messages - Ende Jin

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16
MAT334--Lectures & Home Assignments / Re: 2.5 - Q23
« on: November 20, 2018, 08:24:14 PM »
RE: Min Gyu Woo

If you only want a legitimate procedure for the answer to the question, you can just "instantiate" the proof.
Since
\begin{align*}
    \frac{z + 2}{ (z-2) (z + 1)} & = \frac{4}{3}(\frac{1}{2} \frac{-1}{1 - \frac{z}{2}}) - \frac{1}{3} (\frac{1}{z}\frac{1}{1+\frac{1}{z}}) \\
    &= - \frac{4}{3} \frac{1}{2} (\sum (\frac{z}{2})^n) - \frac{1}{3} \frac{1}{z}(\sum (\frac{1}{z})^n)
\end{align*}

The second line is correct because the convergence relies on the geometric series instead of (12) and (13).

However, it is reasonable to ask if (12) and (13) can be extended to less than or equal to.

The second part of the question can solve similarly.

17
MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« on: November 19, 2018, 11:35:40 PM »
It is at P406, The solution for Q19.
Actually, Yunfei Xia provided a very similar proof with that.

18
MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« on: November 18, 2018, 11:58:09 PM »
I realized that you provided a solution different from the given one back in the book.
But the question doesn't say $\sum a_n(z-z_0)^n = 0$ when $r < |z-z_0| < R$. It just converges. The series converges to zero only if $0 < |z - z_0| < r$.
That is basically my question. The first condition doesn't seem helpful.

19
MAT334--Lectures & Home Assignments / Re: 2.5 Q19
« on: November 18, 2018, 04:58:24 PM »
What is the first condition used for? The one that the series converges in the $\{z: r < |z| < R\}$.

20
Quiz-6 / Re: Q6 TUT 0101
« on: November 17, 2018, 07:21:15 PM »
Thus there exists analytic $g$ s.t. $f(z) = (z-z_0)^mg(z)$ where $g(z_0) \neq 0$.

Thus there exists a small ball around $z_0$ s.t. $g(z) \neq 0$ (by continuity) and analytic ,  which means $\frac{1}{g(z)}$ is analytic as well, thus $\frac{g'(z)}{g(z)}$ is analytic on that ball as well.

Since $m \ge 1$,
\begin{align*}
    \frac{f'(z)}{f(z)} &= \frac{m(z-z_0)^{m-1}g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)} \\
    & = \frac{m(z-z_0)^{m-1}g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)} \\
    &= \frac{g'(z)}{g(z)} + m \frac{1}{z-z_0}
\end{align*}
We have shown $\frac{g'}{g}$ is analytic on that ball. Thus the residue, which means the coefficient of $(z-z_0)^{-1}$ is only $m$ .

21
MAT334--Lectures & Home Assignments / Re: Poles and several singularities
« on: November 10, 2018, 04:46:45 PM »
1. I understand that there can be a small ball around $z_0$ that $f$ are very big, since the limit is infinity. However, for example, if f is analytic on $0 <|z-z_0| < r_0$ and we want to find the decomposition on $0 <|z-z_0| < r_0$ . That means I will have to use the limit first so that for some $r_1$, $|f(z)| > 1 \forall 0 < |z  - z_0| < r_1$. After that, we have a decomposition on $0 < |z  - z_0| < r_1$. But that decomposition is only valid in the $ 0 < |z  - z_0| < r_1$ while we need to find a decomposition on $ 0 < |z  - z_0| < r_0$.

I was thinking about adding a polynomial but I am not so sure because what if after adding polynomial, there is still some zero inside the range?

22
MAT334--Lectures & Home Assignments / Re: Poles and several singularities
« on: November 10, 2018, 12:34:53 PM »
I suddenly realized that I can plus a polynomial (big enough in the domain) in my question 1.

Although that doesn't help with the question 2.

23
MAT334--Lectures & Home Assignments / Poles and several singularities
« on: November 10, 2018, 12:28:42 PM »
1. In the book, when talking about poles: (see attachment 1)
It declares "there is no harm to assume $|f(z)| > 1 $ in $0 < |z - z_0| < r_0$". But why there is no harm? I mean I understand there exists a small ball around $z_0$ such that $f(z)$ can be very big, however, you see that $g(z) = \frac{1}{f(z)}$, if there is a point in $0 < |z_1 - z_0| < r_0$ s.t. $f(z_1) = 0$, that means I must find a smaller $r_0' < |z_1 - z_0| \le r_0$ ,s.t., $|f(z)| > 1 $ in $0 < |z - z_0| < r_0'$ and go on with this proof. However, that means the decomposition $\frac{H(z)}{(z-z_0)^m} = f(z)$ is only valid in $0 < |z - z_0| < r_0'$, then what about the part $r_0' \le |z - z_0| < r_0$?

2. (Attachment 2) I cannot understand this part: what does repeat mean? I have no idea how to extend the above argument into the situation where there are several poles in the domain (). I can understand how to do it when there are only several removable singularities though.

24
In a tutorial, the TA showed us why FTOC II can be used in the complex context, the proof is simply just making the derivative of F into the imaginary part $v'$ and real part $u'$. Because that is a line integral, thus ultimately,  replace the $\Gamma$ with the parameterization of $\gamma : [a, b] \rightarrow \mathbb{C}$, we are integrating the derivative of $u \circ \gamma: [a,b] \rightarrow \mathbb{R}$ which is just a integral in the real line. Now after using the real-function version of FTOC II on the derivative of a real-valued function, we get $u,v$ back and end up with $F$ again.

This is the outline of the proof. However, I don't see that F needs to be analytic on a simply-connected domain in this proof. That F can be analytic on only an annulus (i.e. differentiable on the range of parameterization), even F is undefined outside the annulus, we can still get that the integration is zero when integrating on a closed curve. But it is absolutely wrong because it contradicts the chapter of singularities.

Which part is wrong?

25
Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P2
« on: November 03, 2018, 05:24:10 PM »
we know
\begin{align*}
    \int \sqrt{1 - z^2} dz &= z \sqrt{1 - z^2} - \int z d \sqrt{1-z^2} \\
    & = z \sqrt{1-z^2}  + \int \frac{1}{\sqrt{1 - z^2}}dz - \int \sqrt{1 - z^2} dz
\end{align*}
Thus
\begin{align*}
    2 \int \sqrt{1 - z^2}dz = z\sqrt{1-z^2} + \arcsin z
\end{align*}

Thus
\begin{align*}
    F(z) &= \int f(z^2) \\
    & = \int (\sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!}z^{2n} +2 )dz \\
    & = 2z + \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} \frac{1}{2n+1} z^{2n+1} \\
    & = 2z + \sum_{n=1}^\infty \frac{1}{n!} \frac{1}{2n+1} (-1) \frac{1}{2^{n-1}} \prod_{j=1}^{n-1}(2j-1) z^{2n+1} \\
    &= 2z -  \sum_{n=1}^\infty \frac{1}{n!} \frac{1}{2n+1} \frac{\prod_{j=1}^{n-1}(2j-1) }{2^{n-1}} z^{2n+1}
\end{align*}

26
Quiz-5 / Re: Q5 TUT 5301
« on: November 03, 2018, 02:29:15 PM »
For the former one, the largest disc is $\{z: |z + 1| < 2\}$

For the latter one, the largest disc is the whole complex plane.

27
Quiz-5 / Re: Q5 TUT 5101
« on: November 03, 2018, 01:46:29 PM »
The largest disc is $\{z : |z| < \frac{\pi}{2}\}$

28
Quiz-3 / Re: Q3 TUT 5201
« on: October 31, 2018, 01:18:31 PM »
Can you elaborate that part using the identity of $\sinh, \cosh$? Where to use it?

29
Quiz-3 / Re: Q3 TUT 5201
« on: October 30, 2018, 09:33:50 PM »
Showing Bijective: We first show it is bijective between $\{x +yi : 0 < x < \frac{\pi}{2}, y > 0\}$ and $\{ai + b : a $>$ 0 , b$>$ 0 \} $ and because of symmetry ($\sin(\bar{x}) = \overline{\sin (x)}$ and $\sin(-x) = -\sin(x)$) we can get other quadrant for free (not including axis)\\

{When $a > 0, b > 0$}


Let $z = x+yi$ where $x, y \in \mathbb{R}$
\begin{align*}
    \sin(z) & = \sin(x+yi) \\
    & = i \cos x \frac{e^{y} - e^{-y}}{2} + \sin x \frac{e^{-y} + e^{y}}{2}
\end{align*}

Equivalently, we need to show, for all $a > 0, b > 0$
\begin{align*}
        \cos x \frac{e^{y} - e^{-y}}{2}  &= a \\
     \sin x \frac{e^{-y} + e^{y}}{2} &= b
\end{align*}
Has one and only one solution in $x \in (0, \frac{\pi}{2}), y > 0$.
We make a manipulation that,

from (Equation 1)$^2$ + (Equation 2)$^2$

and (Equation 1)$^2$ - (Equation 2)$^2$ we can get
\begin{align*}
    \frac{e^{2y} + e^{-2y}}{4} - \frac{1}{2} \cos 2x &= a^2 + b^2 \\
    \frac{e^{2y} + e^{-2y}}{4} \cos 2x - \frac{1}{2} &= a^2 - b^2
\end{align*}

Let $u = e^{2y} + e^{-2y}, v = \cos 2x$ (we can see that both $u,v$ are injective when $x \in (0, \frac{\pi}{2}), y > 0$),
 we get equations
 \begin{align}
     \frac{u}{4} - \frac{v}{2} &= a^2 + b^2  \label{eq:3}\\
     \frac{uv}{4} - \frac{1}{2} &= a^2 - b^2 \label{eq:4}
 \end{align}
 Thus, we only need to show $u,v$ has one and only one solution in the above equations where $u \in [2, \infty), v \in [-1,1]$
We can eliminate $u, v$ in \ref{eq:4} respectively by substituting from \ref{eq:3}. Thus we get,
\begin{align}
    v^2 + 2(a^2+b^2)v - (2(a^2-b^2) + 1) &= 0 \\
    u^2 - 4(a^2+b^2)u - (8(a^2-b^2) + 4) &= 0
\end{align}
define 
\begin{align*}
        f(v) &= v^2 + 2(a^2+b^2)v - (2(a^2-b^2) + 1) \\
    g(u) &= u^2 - 4(a^2+b^2)u - (8(a^2-b^2) + 4)
\end{align*}

Since $f(-1) = -4a^2 < 0$ and $f(1) = 4b^2 > 0$ thus by intermediate theorem, there is one  solution for $v$ in (-1,1), and it is the only one in (-1,1) because it is a parabola.

Since $g(2) = -16a^2 < 0$ and $\lim_{x \rightarrow \infty} g(x) = \infty$, again by intermediate theorem, there is one solution for $u$ in (2, $\infty$], and it is the only one in  (2, $\infty$]  because it is a parabola.

{When $a = 0, 0 < b \le 1$}
Similarly as above, we get
\begin{align*}
        \cos x \frac{e^{y} - e^{-y}}{2}  &= 0 \\
     \sin x \frac{e^{-y} + e^{y}}{2} &= b
\end{align*}
Thus, $y = 0$, it is trivial to see there is a solution $\sin x = b$ since $b < 1$, it is the only one because $\sin$ is injective in $(-\frac{\pi}{2}, \frac{\pi}{2})$

{When $a > 0, b = 0$}
\begin{align*}
        \cos x \frac{e^{y} - e^{-y}}{2}  &= a \\
     \sin x \frac{e^{-y} + e^{y}}{2} &= 0
\end{align*}
Thus $x = 0$, since $\frac{e^{y} - e^{-y}}{2}  = a \Leftrightarrow e^{2y} -2ae^{y} - 1$ where $\Delta = 4a^2 + 4 > 0$ thus we have a solution. It is the only solution because $\alpha \mapsto e^{\alpha} - e^{-\alpha}$ is an strictly increasing function (by derivative), which means injective.

We have shown there is one and only one solution in domain from different parts of the codomain. Thus it is bijective

30
MAT334--Lectures & Home Assignments / Section 1.3, Q9 + Q10
« on: September 25, 2018, 11:25:46 AM »
They ask us to show that the sets are open and connected while the solution only shows what the sets are.
Does it mean that in the quiz, we can just show what the set is and then say "thus it is open and connected"? (It's is obvious since it only involves rotation and shrinking and an offset).

Since I think I cannot write the proof down in time in a quiz. Because I have to find an appropriate radius for the ball which involves a complicated computation unless I can use some theorem like:
"a function is continuous if and only if its inverse mapping preserve open" and
In this question, a segment after $z \mapsto \alpha z + \beta$ mapping is a segment which means I can prove polygonal connectedness, while I cannot do the same thing in Q10 though the set is much simpler.

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