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### Messages - Junya Zhang

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1
##### Term Test 2 / Re: TT2 Problem 4
« on: November 24, 2018, 10:26:42 AM »
@liuandon
We are computing a real integral (integral over real values), the result should not have any $i$ in it.

2
##### Term Test 2 / Re: TT2 Problem 4
« on: November 24, 2018, 09:32:01 AM »
Following the hint, consider $f(z)=\frac{1}{\sqrt{z} (z^2 + 1)}$ and contour $\gamma$ as shown in given picture.
Then $$\int_{\gamma}{f(z)dz} =\int^{-\epsilon}_{-R}{f(x)dx} + \int^{R}_{\epsilon}{f(x)dx} + \int_{\gamma_{\epsilon}}{f(z)dz} + \int_{\gamma_{R}}{f(z)dz}$$

Step 1
Fix $0 < \epsilon < 1$ and $R>1$.
Then $f$ is analytic everywhere in $\gamma$ except at $z = i$, where it has a simple pole.
Then by the residue theorem, $$\int_{\gamma}{f(z)dz} = 2\pi i \left[\frac{1}{\sqrt{z}(z+i)}\right]_{z=i} = \frac{2\pi i}{2i\sqrt{i}} = \frac{\pi }{\sqrt{i}}$$ $$= \frac{\pi }{\sqrt{e^{i\pi/2}}} = \frac{\pi }{e^{i\pi/4}} =\frac{\pi }{\cos(\pi/4) + i \sin(\pi/4)} = \frac{\pi }{\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}} = \frac{\pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)} = \frac{\pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)}{\frac{1}{2}+ \frac{1}{2}} = \pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)$$

Overcomplicated, but correct. V.I
Note that this holds for all $0 < \epsilon < 1$ and $R>1$.

Step 2
$$\int^{-\epsilon}_{-R}{f(x)dx} = \int^{-\epsilon}_{-R}{\frac{1}{\sqrt{x} (x^2 + 1)}dx} = \int^{\epsilon}_{R}{\frac{1}{\sqrt{-x} (x^2 + 1)} (-dx)} = \int^{R}_{\epsilon}{\frac{1}{\sqrt{-x} (x^2 + 1)}dx} = \frac{1}{i} \int^{R}_{\epsilon}{\frac{1}{\sqrt{x} (x^2 + 1)}dx}$$

Step 3
$$\int^{R}_{\epsilon}{f(x)dx} = \int^{R}_{\epsilon}{\frac{1}{\sqrt{x} (x^2 + 1)}dx}$$

Step 4
Consider $\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert$. With the parametrization $z = \epsilon e^{i\theta}$ for $\theta\in[0,\pi]$.
$$\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert = \left\lvert{\int_{0}^{\pi}{\frac{i\epsilon e^{i\theta}}{\sqrt{\epsilon e^{i\theta}} ((\epsilon e^{i\theta})^2 + 1)}d\theta}}\right\lvert \leq {\int_{0}^{\pi}{ \frac{\left\lvert i\epsilon e^{i\theta}\right\lvert} {\left\lvert\sqrt{\epsilon e^{i\theta}}\right\lvert \left\lvert (\epsilon e^{i\theta})^2 + 1\right\lvert} d\theta}} = \int_{0}^{\pi} \frac{\epsilon}{\sqrt{\epsilon} \left\lvert (\epsilon e^{i\theta})^2 + 1\right\lvert}d\theta \leq \int_{0}^{\pi} \frac{\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}d\theta = \frac{\pi\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}$$
$$\lim_{\epsilon \to 0^+}{\frac{\pi\epsilon}{\sqrt{\epsilon} (1 - \epsilon^2)}} = \lim_{\epsilon \to 0^+}{\frac{\pi\sqrt{\epsilon}}{ (1 - \epsilon^2)}} = \frac{0}{1} = 0$$
This implies that $$\lim_{\epsilon \to 0^+}{\left\lvert{\int_{-\gamma_{\epsilon}}{f(z)dz}}\right\lvert} = 0$$
Which then gives $$\lim_{\epsilon \to 0^+}{{\int_{-\gamma_{\epsilon}}{f(z)dz}}} = 0$$ and thus $$\lim_{\epsilon \to 0^+}{{\int_{\gamma_{\epsilon}}{f(z)dz}}} = -0 = 0$$

Step 5
Consider $\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert$. With the parametrization $z = R e^{i\theta}$ for $\theta\in[0,\pi]$.
$$\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert = \left\lvert{\int_{0}^{\pi}{\frac{iR e^{i\theta}}{\sqrt{R e^{i\theta}} ((R e^{i\theta})^2 + 1)}d\theta}}\right\lvert \leq {\int_{0}^{\pi}{ \frac{\left\lvert iR e^{i\theta}\right\lvert} {\left\lvert\sqrt{R e^{i\theta}}\right\lvert \left\lvert (R e^{i\theta})^2 + 1\right\lvert} d\theta}} = \int_{0}^{\pi} \frac{R}{\sqrt{R} \left\lvert (R e^{i\theta})^2 + 1\right\lvert}d\theta \leq \int_{0}^{\pi} \frac{R}{\sqrt{R} (R^2-1)}d\theta = \frac{\pi R}{\sqrt{R} (R^2-1)}$$
$$\lim_{R \to \infty}{\frac{\pi R}{\sqrt{R} (R^2 - 1)}} = \lim_{R \to \infty}{\frac{\pi}{\sqrt{R} (R - \frac{1}{R})}} = \frac{\pi}{\infty \cdot \infty} = 0$$
This implies that $$\lim_{R \to \infty}{\left\lvert{\int_{\gamma_{R}}{f(z)dz}}\right\lvert} = 0$$
Which then gives $$\lim_{R \to \infty}{{\int_{\gamma_{R}}{f(z)dz}}} = 0$$

Altogether, as $\epsilon \to 0^+$ and $R \to \infty$, the equation $\int_{\gamma}{f(z)dz} =\int^{-\epsilon}_{-R}{f(x)dx} + \int^{R}_{\epsilon}{f(x)dx} + \int_{\gamma_{\epsilon}}{f(z)dz} + \int_{\gamma_{R}}{f(z)dz}$ becomes
$$\pi \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right) = \frac{1}{i} I + I + 0 + 0$$
Equating the real part of the equation gives $$\frac{\pi}{\sqrt{2}} = I$$
Checking: equating the imaginary part of the equation gives $$-\frac{i\pi}{\sqrt{2}} = \frac{1}{i}I$$
which gives $$I = \frac{\pi}{\sqrt{2}}$$ as well.

3
##### Term Test 2 / Re: TT2A Problem 3
« on: November 24, 2018, 08:55:19 AM »
$$f(z)=(z^2 - 1) \cot(\pi z^2)=(z+1)(z-1)\frac{\cos \pi z^2}{\sin \pi z^2}$$

$f$ has a singularity at $z$ iff $\sin \pi z^2 = 0$, this happens when $z^2 = k$ for $k \in \mathbb{Z}$.
Note that $\cos \pi z^2 = 0$ iff $z^2 = \frac{1}{2} + \mathbb{Z}$, which is mutually exclusive from zeros of $\sin \pi z^2 = 0$.

Case 1: $z = -1$
Since $(z+1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=-1$, then $f$ has a removable singularity at $z=-1$.

Case 2: $z = 0$
Since $\sin \pi z^2 = 0$ has a zero of order 2, and no other factors have zeros at $z=0$, then $f$ has a pole of order 2 at $z=0$ .

Case 3: $z = 1$
Since $(z-1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=1$, then $f$ has a removable singularity at $z=1$.

Case 4: $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$
Since $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at these values, then $f$ has a simple pole at $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$.

Case 5: singularity at $\infty$
Let $$g(w) = f(1/w) = \left(\frac{1}{w}+1\right)\left(\frac{1}{w}-1\right)\frac{\cos \left( \frac{\pi}{w^2}\right)}{\sin \left(\frac{\pi}{w^2}\right)}$$

Note that $g$ has singularities at $w=0$ and $w = \pm \sqrt\frac{1}{k}$ for $k \in \mathbb{Z}$. This implies that $\forall R>0, \exists w\in\mathbb{C}, 0<|w|<R$ such that $g$ is has a singularity at $w$, and thus $g$ is not analytic in $0<|w|<R$ for any $R>0$.
Therefore, the singularity of $g$ at $z=0$ is a non-isolated singularity. And thus, the singularity of $f$ at $\infty$ is a non-isolated singularity.

4
##### Quiz-6 / Re: Q6 TUT 0201
« on: November 17, 2018, 04:20:00 PM »
This is question 8 from CH2.5 in the textbook.

Let $w = z - 1$. Then $z = w+ 1$.
$$\frac{z^2}{z^2 -1} = \frac{(w+1)^2}{(w+1)^2 -1} = \frac{w^2 + 2w + 1}{w^2 + 2w}= 1+ \frac{1}{w^2 + 2w}=1 + \frac{1}{2w} \cdot \frac{1}{1 + w/2} = 1 + \frac{1}{2w} \sum_{n = 0}^{\infty} \left(- \frac{w}{2}\right)^n$$ which is valid for $|w/2| < 1$, i.e. $|z-1|<2$. Then
$$\frac{z^2}{z^2 -1} = 1 + \frac{1}{2w}\left(1 - \frac{w}{2} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n}w^n \right) = 1 + \frac{1}{2w} - \frac{1}{4} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n-1}w^{n-1} = \frac{1}{2w} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} 2^{-n-2}w^{n} = \frac{1}{2w} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{w^{n}}{2^{n+2}}$$ Substitute $w = z-1$ back to the equation we get $$\frac{z^2}{z^2 -1} = \frac{1}{2(z-1)} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{(z-1)^{n}}{2^{n+2}}$$
Residue of the function at $z = 1$ is the coefficient of $\frac{1}{z-1}$ in the Laurent series, which is $\frac{1}{2}$.

5
##### Quiz-5 / Re: Q5 TUT 0201
« on: November 02, 2018, 10:45:59 PM »
Morera's Theorem:
If $f$ is a continuous function on a domain $D$, and if $\int_{\gamma} f(z) dz = 0$ for every triangle $\gamma$ that lies, together with its interior, in $D$, then $f$ is analytic on $D$.

Define $f(z) = \int_0^1 \frac{dt}{1-tz}$ and let $D = \{z\in\mathbb{C}:|z|≤1\}$
Let $\gamma$ be a triangle that lies together with its interior in $D$.
Note that $1-tz = 0$ iff $z = \frac{1}{t}$ which implies $|z| = |1/t| ≥ 1$ (since $t\in[0,1]$)
Therefore, we can conclude that $1-tz ≠ 0$ for all $z \in \{z\in\mathbb{C}:|z|<1\}$.
This implies that $f$ is continuous on $\{z\in\mathbb{C}:|z|<1\}$, and for fixed $t\in[0,1]$, $f$ is analytic on and inside $\gamma$.
By Cauchy's theorem, $$\int_\gamma \frac{1}{1-tz} dz = 0$$ for fixed $t\in[0,1]$
Then, $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( \int_0^1 \frac{dt}{1-tz}\right) dz = \int_0^1 \left(\int_{\gamma}\frac{1}{1-tz} dz\right) dt = \int_0^1 0 dt = 0$$
By Morera's Theorem, we can conclude that $\int_0^1 \frac{dt}{1-tz}$ is analytic on the domain $|z|<1$.

$$f(z) = \int_0^1 \frac{dt}{1-tz} = \int_0^1 \sum_{n=0} ^{\infty} (tz)^n dt = \sum_{n=0} ^{\infty} \int_0^1(tz)^n dt = \sum_{n=0} ^{\infty} \left[\frac{z^n\cdot t^{n+1}}{n+1}\right]_{0}^{1} = \sum_{n=0} ^{\infty} \frac{z^n}{n+1}$$

6
##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P1
« on: October 30, 2018, 12:40:19 PM »
Deformation Theorem:
Let $\gamma$ be a simple closed curve. Let $D(a, r)$ be an open disc lies inside $\gamma$.
Let $f$ be an analytic function on and inside $\gamma$ except maybe at $a$.
Then $\int_{\gamma} f(z) dz = \int_{|z-a|=r} f(z) dz$

In our question, $f(z) = \frac{1}{z^3 - 1}$, which is analytic everywhere except at the three points labeled on the graph, namely, 1, $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Let A, B, C be three complex numbers such that $f(z) = \frac{A}{z-1} + \frac{B}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{C}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$
Solve for A,B,C we get
$$A = \frac{1}{3}$$ $$B = \frac{2}{-3\sqrt{3}i-3}$$ $$C = \frac{2}{3\sqrt{3}i-3}$$
So, $f(z) = \frac{\frac{1}{3}}{z-1} + \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$

Let $\gamma$ be the blue curve.
The Deformation theorem implies that $$\int_{\gamma} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{1}{z^3 - 1} dz$$ for some $r>0$.

$$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz =\int_{|z-1|=r} \frac{\frac{1}{3}}{z-1} + \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz = \int_{|z-1|=r}\frac{\frac{1}{3}}{z-1} dz +\int_{|z-1|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz + \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz$$
Since $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$ are outside of the unit circle $|z-1|=r$, then $\int_{|z-1|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz = 0$ and $\int_{|z-1|=r} \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz =0$ by Cauchy's theorem.
Therefore, $\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{\frac{1}{3}}{z-1} dz = \frac{1}{3}\int_{|z-1|=r} \frac{1}{z-1} dz = \frac{1}{3}\cdot 2\pi i = \frac{2\pi i}{3}$.
Thus, the integral of $f$ on the blue curve is $\frac{2\pi i}{3}$.

Similarly, for the pink curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz =\int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz = \frac{2}{-3\sqrt{3}i-3} \cdot 2\pi i = \frac{4\pi i}{-3\sqrt{3}i-3}$

For the brown curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz = \frac{2}{3\sqrt{3}i-3} \cdot 2 \pi i = \frac{4 \pi i}{3\sqrt{3}i-3}$

For the green curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4\pi i}{-3\sqrt{3}i-3}$

For the orange curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$

For the red curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4 \pi i}{3\sqrt{3}i-3}$

For the purple curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$

This concludes the question.

7
##### Quiz-3 / Re: Q3 TUT 0201
« on: October 12, 2018, 06:26:46 PM »
This is question 23 from textbook section 1.5.

(1) Show that $F(z) = e^z$ maps the strip $S=\{x+iy:-\infty < x < \infty, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\}$ onto the region $\Omega = \{w=s+it:s\geq 0, w \neq 0\}$.
Let $z=x+iy$ be a complex number in $S$. Then we know that $x\in(-\infty, \infty)$ and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$$F(z) = e^{x+iy} = e^xe^{iy} = e^x(\cos y + i\sin x) = e^x\cos y + ie^x\sin y$$
Let $F(z)=w=s + it$. Then $s = e^x\cos y$ and $t = e^x\sin y$.
Note that $\forall y \in [-\frac{\pi}{2}, \frac{\pi}{2}], \cos y \geq 0$ and $\forall x \in (-\infty, \infty), e^x > 0$, thus $s = e^x\cos y \geq 0$.
Moreover, $s = 0$ if and only if $\cos y = 0$, which happens when $y = \frac{\pi}{2}$ or $-\frac{\pi}{2}$. However, when $y = \frac{\pi}{2}$, $t = e^x\sin y = e^x \neq 0$ and when $y = -\frac{\pi}{2}$, $t = e^x\sin y = -e^x \neq 0$. This shows that $F(z) = w = s+it \neq 0$.
Thus, $F(z)\in \Omega$.
Since $z$ was an arbitrary element in $S$, this shows that $F$ maps the strip $S$ onto the region $\Omega$.

(2) Show that the $F$ is one-to-one on $S$.
To show that $F$ is one-to-one on $S$, we need to show that $\forall z_1, z_2 \in S, F(z_1)=F(z_2)\Rightarrow z_1=z_2$.
Let $z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$ be two arbitrary elements in $S$, and suppose $F(z_1)=F(z_2)$. Then by computation in part (1) we have $$F(z_1) = e^{x_1}\cos {y_1} + ie^{x_1}\sin {y_1}$$ $$F(z_2) = e^{x_2}\cos {y_2} + ie^{x_2}\sin {y_2}$$
Equating $F(z_1)$ and $F(z_2)$ we get $$e^{x_1}\cos {y_1} = e^{x_2}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_2}\sin {y_2}$$
Square both sides of the two equations we get $$e^{2x_1}\cos^2 {y_1} = e^{2x_2}\cos^2 {y_2}$$ $$e^{2x_1}\sin^2 {y_1} = e^{2x_2}\sin^2 {y_2}$$
Add the two equations together we get $$e^{2x_1}(\cos^2 {y_1} + \sin^2 {y_1}) = e^{2x_2}(\cos^2 {y_2} + \sin^2 {y_2} )$$
Simplify this we get $e^{2x_1} = e^{2x_2}$. Since $e^{2x_2} \neq 0$, we can divide both sides of the equation by $e^{2x_2}$ and get $e^{2x_1 - 2x_2} = 1$, which implies $2x_1 - 2 x_2 = 0$. Therefore we can conclude $x_1 = x_2$.

Back to the equations $$e^{x_1}\cos {y_1} = e^{x_2}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_2}\sin {y_2}$$
Now substitute in $x_1 = x_2$ we have $$e^{x_1}\cos {y_1} = e^{x_1}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_1}\sin {y_2}$$
This implies $$\cos {y_1} = \cos {y_2}$$ $$\sin {y_1} = \sin {y_2}$$
Multiply the first equation by $\sin {y_2}$ and the second equation by $\cos {y_2}$ we get $$\sin {y_2}\cos {y_1} = \sin {y_2}\cos {y_2}$$ $$\sin {y_1}\cos {y_2} = \sin {y_2}\cos {y_2}$$
Subtract the second equation from the first we get $$\sin {y_2}\cos {y_1} - \sin {y_1}\cos {y_2} = \sin {y_2}\cos {y_2} - \sin {y_2}\cos {y_2}$$
Simplify this we get $$\sin ({y_2}- {y_1})= 0$$
This implies ${y_2}- {y_1} = k\pi$ for $k\in\mathbb{Z}$.
Since $y_1,y_2\in [-\frac{\pi}{2}, \frac{\pi}{2}]$, then $y_2 - y_1 \in [-\pi,\pi]$.
Together, possible values for $y_2 - y_1$ are $-\pi, 0, \pi$.
If $y_2 - y_1 =-\pi$, then $y_2 = -\frac{\pi}{2}$ and $y_1 = \frac{\pi}{2}$.
If $y_2 - y_1 =\pi$, then $y_1 = -\frac{\pi}{2}$ and $y_2 = \frac{\pi}{2}$.
From $\sin {y_1} =\sin {y_2}$ we can eliminate these two cases since $\sin(-\pi/2) = -1$ and $\sin(\pi/2) = 1$
Therefore, it must be the case that $y_1 - y_2 = 0$. That is $y_1 = y_2$.

Since $x_1 = x_2$ and $y_1 = y_2$, we can conclude that $z_1 = z_2$. This shows that $F$ is one-to-one on $S$.

(3) Show that $F$ maps the boundary of $S$ onto all the boundary of $\Omega$ except $w=0$.
The boundary of $S$ is the set $\{x+iy : -\infty < x < \infty, y = \frac{\pi}{2} \text{ or } y = -\frac{\pi}{2}\} = \{z: \text{Im }z = \frac{\pi}{2}\} \cup \{z: \text{Im }z = \frac{-\pi}{2}\}$ and the boundary of $\Omega$ is the set $\{s+it:s = 0\}$
We will show in part (4) that $F$ maps the set $\{z: \text{Im }z = \frac{\pi}{2}\}$ to $\{s+it:s = 0, t > 0\}$ and maps the set $\{z: \text{Im }z = \frac{-\pi}{2}\}$ to $\{s+it:s = 0, t < 0\}$, which we can then conclude that $F$ maps the boundary of $S$ onto the boundary of $\Omega$ except for the origin.

(4) Explain what happens to each of the horizontal lines $\{z: \text{Im }z = \frac{\pi}{2}\}$ and $\{z: \text{Im }z = -\frac{\pi}{2}\}$.
Let $z=x+i\pi/2$ be a complex number on the line $\{z: \text{Im }z = \frac{\pi}{2}\}$. The function $F(z)=e^z$ maps $z$ to $e^x\cos \pi/2 + ie^x\sin \pi/2 = ie^x$. Since the function $f(x) = e^x$ is a one-to-one and onto function from $(-\infty, \infty)$ to $(0,\infty)$, this shows that $F$ maps the line $\{z: \text{Im }z = \frac{\pi}{2}\}$ onto the positive imaginary axis (not including the origin).
Similarly, let $z=x+i(-\pi/2)$ be a complex number on the line $\{z: \text{Im }z = \frac{-\pi}{2}\}$. The function $F(z)=e^z$ maps $z$ to $e^x\cos (-\pi/2) + ie^x\sin (-\pi/2) = -ie^x$. Since $f(x) = -e^x$ is a one-to-one and onto function from $(-\infty, \infty)$ to $(-\infty, 0)$, this shows that $F$ maps the line $\{z: \text{Im }z = -\frac{\pi}{2}\}$ onto the negative imaginary axis (not including the origin).

(5) Draw both domains.
See image from textbook page 55.

8
##### Thanksgiving bonus / Re: Thanksgiving bonus 3
« on: October 06, 2018, 12:30:41 PM »
Let $f(x,y) = u+iv = e^y\cos(x)+ie^y\sin(x)$.
Then
$$\bar{f}(z,y) = e^y\cos(x)-ie^y\sin(x)$$ $$u(x,y)=e^y\cos(x)$$ $$v(x,y)=e^y\sin(x)$$
The given function represents a locally sourceless and irrotational flow since $\bar{f}$ is analytic on $\mathbb{C}$.

Note that
$$\frac{\partial{v}}{\partial{x}} = e^y\cos(x)$$ $$\frac{\partial{u}}{\partial{y}} = e^y\cos(x)$$ $$\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}} = 0$$
This shows that $f$ is globally irrotational.
$$\frac{\partial{u}}{\partial{x}} = -e^y\sin(x)$$ $$\frac{\partial{v}}{\partial{y}} = e^y\sin(x)$$ $$\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0$$
This shows that $f$ is globally sourceless.
So $f$ is both globally sourceless and globally irrotational.

9
##### Thanksgiving bonus / Re: Thanksgiving bonus 2
« on: October 06, 2018, 11:58:38 AM »
Let $f(x,y) = u+iv = x^2-y^2 + 2ixy$.
Then
$$\bar{f}(z,y) = x^2 - y ^2 -2ixy$$ $$u(x,y)=x^2-y^2$$ $$v(x,y)=2xy$$
The given function represents a locally sourceless and irrotational flow since $\bar{f}$ is analytic on $\mathbb{C}$.
However, $f$ is neither globally sourceless nor globally irrotational.
$$\frac{\partial{v}}{\partial{x}} = 2y$$ $$\frac{\partial{u}}{\partial{y}} = -2y$$ $$\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}} = 4y$$
This shows that $f$ is not globally irrotational.
$$\frac{\partial{u}}{\partial{x}} = 2x$$ $$\frac{\partial{v}}{\partial{y}} = 2x$$ $$\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 4x$$
This shows that $f$ is not globally sourceless.

10
##### Quiz-2 / Re: Q2 TUT 5301
« on: October 06, 2018, 10:37:45 AM »
The limit does not exist.
By definition of limit as $z\to\infty$,
$$\lim_{z\to\infty} h(z) =\lim_{z\to\infty} \frac{|z|}{z} = \lim_{z\to 0} \frac{|\frac{1}{z}|}{\frac{1}{z}} =\lim_{z\to 0} \frac{\frac{1}{|z|}}{\frac{1}{z}} = \lim_{z\to 0} \frac{z}{|z|}$$
Let $z = x + iy$, then
$$\lim_{z\to\infty} h(z) = \lim_{(x,y)\to (0,0)} \frac{x+iy}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} + i\frac{y}{\sqrt{x^2+y^2}}$$
Note that $\lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}}$ does not exist since $$\lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} = 1$$ when $z$ approaches 0 alone the positive real axis, and $$\lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} = -1$$ when $z$ approaches 0 alone the negative real axis.

Similarly, $\lim_{(x,y)\to (0,0)} \frac{y}{\sqrt{x^2+y^2}}$ does not exist.
This implies that $\lim_{z\to\infty} h(z)$ does not exist.

11
##### Quiz-1 / Re: Q1: TUT 0201 and TU 0202
« on: September 28, 2018, 06:00:01 PM »
Let $z=x+iy$ denote an arbitrary point on the perpendicular bisector. Then we have $$|z-(-1+2i)| = |z-(1-2i)|$$
Simplify this we get $$|(x+1)+i(y-2)| = |(x-1) + i(y+2)|$$
Square both sides we have $$|(x+1)+i(y-2)|^2 = |(x-1) + i(y+2)|^2$$
Which is $$(x+1)^2+(y-2)^2=(x-1)^2+(y+2)^2$$
Expand the squares we get $$x^2+2x+1+y^2-4y+4=x^2-2x+1+y^2+4y+4$$
Cancel same terms from both sides we have $$4x=8y$$
That is $$y=\frac{1}{2}x$$
According to the formula given on page 13 of the textbook, equation of this perpendicular bisector is $$Re\left(\left(\frac{1}{2}+i\right)z\right) = 0$$ where $z\in\mathbb{C}$

12
##### Quiz-6 / Re: Q6--T5101
« on: March 18, 2018, 11:56:12 AM »
Cheng Sheng's solution is correct, but here's the typed solution a)
Let $$P=\begin{pmatrix} -1 & -4 \\ 1 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = \det\begin{pmatrix} -1-\lambda & -4 \\ 1 & -1-\lambda\end{pmatrix}=(\lambda +1)^2 +4$$
Thus, $$\lambda_1 = -1+2i, \lambda_2 = -1-2i$$
Consider $\lambda_1 =-1+2i$.
$$N(P- (-1+2i)I) = N\begin{pmatrix} -1+1-2i & -4 \\ 1 & -1+1-2i\end{pmatrix} = N\begin{pmatrix} -2i & -4 \\ 1 & -2i\end{pmatrix}= span\{\begin{pmatrix} 2i \\ 1\end{pmatrix}\}$$
Consider $$e^{(-1+2i)t}\begin{pmatrix} 2i \\ 1\end{pmatrix} =e^{-t} (\cos(2t) + i\sin(2t))\begin{pmatrix} 2i \\ 1\end{pmatrix}\ = e^{-t}\begin{pmatrix} 2i\cos(2t) -2\sin(2t) \\ \cos(2t) + i\sin(2t) \end{pmatrix} = e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + i e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1 e^{-t}\begin{pmatrix} -2\sin(2t) \\ \cos(2t) \end{pmatrix} + c_2 e^{-t}\begin{pmatrix} 2\cos(2t) \\ \sin(2t) \end{pmatrix}$$

b)
As $t\to\infty$, solution asymptotically converges to the origin in counter clockwise direction in spiral shapes.
See attached image.

13
##### Quiz-6 / Re: Q6--T0101
« on: March 18, 2018, 11:37:36 AM »
Cheng Sheng's solution is correct, but here's the typed solution a)
Let $$P=\begin{pmatrix} 2 & -5 \\ 1 & -2\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = det\begin{pmatrix} 2-\lambda & -5 \\ 1 & -2-\lambda\end{pmatrix}=\lambda^2 + 1$$
Thus, $$\lambda_1 = i, \lambda_2 = -i$$
Consider $\lambda_1 = i$.
$$N(P-iI) = N\begin{pmatrix} 2-i & -5 \\ 1 & -2-i\end{pmatrix} = span\{\begin{pmatrix} 2+i \\ 1\end{pmatrix}\}$$
Consider $$e^{it}\begin{pmatrix} 2+i \\ 1\end{pmatrix} = \cos(t) + i\sin(t)\begin{pmatrix} 2+i \\ 1\end{pmatrix}\ = \begin{pmatrix} 2\cos(t) +2i\sin(t)+i\cos(t) - \sin(t) \\ \cos(t) + i\sin(t) \end{pmatrix} =\begin{pmatrix} 2\cos(t) - \sin(t) \\ \cos(t) \end{pmatrix} + i \begin{pmatrix} 2\sin(t) + \cos(t) \\ \sin(t)\end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1\begin{pmatrix} 2\cos(t) - \sin(t) \\ \cos(t) \end{pmatrix} + c_2 \begin{pmatrix} 2\sin(t) + \cos(t) \\ \sin(t)\end{pmatrix}$$

b)
As $t\to\infty$, solution circulates in counter clockwise direction in elliptical shapes.
See attached image.

14
##### Quiz-6 / Re: Q6--T0701
« on: March 18, 2018, 11:20:29 AM »
Cheng Sheng's solution is correct, but here's the typed solution a)
Let $$P=\begin{pmatrix} 1 & 2 \\ -5 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = det\begin{pmatrix} 1-\lambda & 2 \\ -5 & -1-\lambda\end{pmatrix}=\lambda^2 + 9$$
Thus, $$\lambda_1 = 3i, \lambda_2 = -3i$$
Consider $\lambda_1 = 3i$.
$$N(P-3iI) = N\begin{pmatrix} 1-3i & 2 \\ -5 & -1-3i\end{pmatrix} = span\{\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\}$$
Consider $$e^{3it}\begin{pmatrix} -2 \\ 1-3i\end{pmatrix} = \cos(3t) + i\sin(3t)\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\ = \begin{pmatrix} -2\cos(3t) -2i\sin(3t) \\ \cos(3t) + i\sin(3t) - 3i\cos(3t) + 3\sin(3t)\end{pmatrix} =\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + i \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + c_2 \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$

b)
As $t\to\infty$, solution circulates in clockwise direction in elliptical shapes.
See attached image.

15
##### Quiz-5 / Re: Q5--T0401, T0901
« on: March 09, 2018, 06:19:35 PM »
Junjie's solution is correct, but here's a more detailed version a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{1}{2}x_1''$$
Substitute into the second equation and simplify, we get $$x_1'' + 4 x_1 = 0$$
which is a second order ODE of $x_1$.

b)
Characteristic equation is $r^2 +4 = 0$ with roots $r_1 = 2i, r_2 = -2i$
General solution for $x_1$ is $x_1 = c_1 \cos{2t} + c_2 \sin{2t}$
Plug in to $x_2 = \frac{1}{2}x_1'$ get
$$x_2 = -c_1 \sin{2t} + c_2 \cos{2t}$$
So, $$x_1 = c_1 \cos{2t} + c_2 \sin{2t}$$ $$x_2 = -c_1 \sin{2t} + c_2 \cos{2t}$$
Plug in $x_1(0)=3, x_2(0) = 4$ to get
$$c_1 = 3, c_2 = 4$$
That is, $$x_1 = 3 \cos{2t} + 4 \sin{2t}$$ $$x_2 = -3 \sin{2t} + 4 \cos{2t}$$

c) See attached graph.
Note that the graph is a circle center at origin with radius 5, and as $t\to \infty$, it is moving clockwise.

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