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### Messages - Ende Jin

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1
##### 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".

2
##### 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.

3
##### 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.

4
##### 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 $f$.

5
##### 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 $m$ .

6
##### 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*}

7
##### 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.

8
##### Quiz-5 / Re: Q5 TUT 5101
« on: November 03, 2018, 01:46:29 PM »
The largest disc is $\{z : |z| < \frac{\pi}{2}\}$

9
##### 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?

10
##### 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

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