Toronto Math Forum

@liuandon
We are computing a real integral (integral over real values), the result should not have any $i$ in it.

$$f(z)=(z^2 - 1) \cot(\pi z^2)=(z+1)(z-1)\frac{\cos \pi z^2}{\sin \pi z^2}$$

$f$ 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.

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 $|z|<1$.

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

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.

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.

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 $z\in\mathbb{C}$

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.

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.