### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Topics - RunboZhang

Pages: 1 [2]
16
##### Quiz 4 / Quiz-5101-C
« on: October 23, 2020, 03:48:21 PM »
$\textbf {Problem:} \\\\$
$\text{Evaluate the given integral using the technique of Example 10 of Section 2.3:} \\$
$\begin{gather} \int_{\gamma} e^{z}\, dz \end{gather}$
$\text{where}\ \gamma \ \text{is the semicircle from -1 to 1 passing through i.}$

$\textbf{Solution: } \\\\$
$\text{We have integrand} f(z) = e^{z} \text{, and it is the derivative of }F(z)=e^{z} . \\\\$
$\text{This is valid when } F(z) \text{is analytic on domain D.}\\\\$
$\text{Indeed, both} F(z) \text{ and } f(z) \text{ is analytic on the semicircle.}\\\\$
$\text{Therefore, we have}\\\\$

\begin{gather} \begin{aligned} \int_{\gamma} e^{z}\, dz &{} = \int_{\gamma} f(z)\, dz \\\\ &{} = \int_{\gamma} F'(z)\, dz \\\\ &{} = \text{F(endpoint) \m F(initialpoint)} \\\\ &{} = F(1) - F(-1) \\\\ &{} = e - e^{-1} \end{aligned} \end{gather}

$\text{Therefore} \ e - e^{-1} \ \text{is our final answer.}$

17
##### Quiz 3 / LEC 5101 Quiz3 Question 5A
« on: October 09, 2020, 10:06:44 PM »
Problem:
Show that $F(z) = e^{z}$ maps the strip $S = \{x+iy: -\infty < x < \infty, -\pi/2 \leq y \leq \pi/2\}$ onto the region $D = \{w = s + it: s \geq 0, w \ne 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary of $Q$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.

Proof:
Firstly, we are going to expand the equation and prove that $F$ maps the strip $S$ onto the region $D$.
Let $F(z) = z_0 = u_0 + iv_0$.
$F(z) = e^{z} = e^{x+iy} = e^{x} \cdot e^{iy} = e^{x} \cdot (cos(y) + isin(y)) = e^{x}cos(y) + ie^{x}sin(y) = u_0 + iv_0$.
Note $x \in \mathbb{R}, y \in [-\pi/2, \pi/2]$, we may conclude that $e^{x} \ge 0, cos(y) \in [0,1], and \sin(y) \in [-1,1]$.
Then we have $Re(z_0) = u_0 = e^{x}cos(y) \ge 0$.
Although we get $Re(z_0) = u_0 = 0$ when $x = \pi/2$ or $-\pi/2, Im(z_0) \ne 0$. Thus after mapping, $z_0$ takes all positive x-axis except origin.
Moreover, we have $Im(z_0) = v_0 = e^{x}sin(y) \in \mathbb{R}$.
Since $z_0$ is an arbitrary element we chose from S, we have proved S gets mapped onto D.

Secondly, we are proving $F(z)$ is one-to-one mapping on S.
Equivalently, WTS $F(z_1) = F(z_2) \Longrightarrow z_1 = z_2$.
$\begin{gather} F(z_1) = F(z_2) \\ e^{x_1}\cdot e^{iy_1} = e^{x_2} \cdot e^{iy_2} \\ {\frac{e^{x_1} \cdot e^{iy_1}}{e^{x_2}\cdot e^{iy_2}}} = 1 \\ e^{x_1-x_2} \cdot e^{i(y_1-y_2)} = 1 \end{gather}$
That implies $Re(z_1) = Re(z_2)$ and $Im(z_1) = Im(z_2)$, alternatively, $x_1 = x_2$ and $y_1 = y_2$
Therefore we have proved F is one-to-one on S.

Lastly, we will discuss what happens to the boundary $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.
When $y = \frac {\pi}{2}$:
$\begin{gather} \begin{split} F(x + i\pi/2) & = e^{x} \cdot e^{\frac {i\pi}{2}} \\ & = e^{x} \cdot (cos(\pi/2) + isin(\pi/2)) \\ & = ie^{x} \end{split} \end{gather}$

When $y = \frac {-i\pi}{2}$:
$\begin{gather} \begin{split} F(x - i\pi/2) & = e^{x} \cdot e^{\frac {-i\pi}{2}} \\ & = e^{x} \cdot (cos(-\pi/2) + isin(-\pi/2)) \\ & = -ie^{x} \end{split} \end{gather}$

Since $e^{x}$ > 0, boundaries $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$ is mapped to positive y-axis and negative y-axis except the origin 0.

The pic below is the domain before and after mapping.

18
##### Chapter 2 / Question on 2.1 Example 10
« on: October 02, 2020, 08:08:28 PM »
Hi guys, I just went through chapter 2.1 and found that the example 10 is quite confusing. I am particularly wondering, firstly, why the derivative is du/dx+i dv/dx? Why can't we take the derivative with respect to y? It says Theorem 3 implies the rationale of taking the derivative. Indeed it does, but I could not find how does Theorem 3 indicate the computation of the derivative. Furthermore, by Cauchy-Riemann equation we have du/dx=dv/dy, and du/dy=-dv/dx, if we do take the derivative with respect to y, then the derivative of log(z) would have real part of z and imaginary part of z switched in numerator. Does this cause a problem?

And secondly, it has drawn the conclusion that function log(z) is not analytic on any domain D that contains a simple closed curve that surrounds the origin. So I think log(z) on complex plane is similar as log(x) on real plane, and they both does not have derivative at 0. I have a rough idea about its reasoning but I dont know if I am on the correct track, correct me if I am wrong pls.

Btw I have highlighted the two parts that I have mentioned in my question in the picture below.

19
##### Quiz 2 / LEC0201 Quiz2 Question2B
« on: October 02, 2020, 01:46:06 PM »
Question and my answer is in the pic attached below.

20
##### Chapter 2 / Question on Week3 Lec3 Integrating Factor
« on: September 25, 2020, 04:27:01 PM »
Hi guys, we went through integrating factors in lecture 3 of this week. When using integrating factor to make a non-exact equation exact, we have different types of it. Namely, we have mu(x,y)=mu(x), mu(x,y)=mu(y), and mu(x,y)=mu(xy). I am particularly confused by them. When we are solving a non-exact equation, shall we try three different types and see which one would be helpful, or is there a way to give an instant intuition?

21
##### Chapter 1 / Question on Textbook Section1.5 The Log Function Example 3
« on: September 22, 2020, 09:16:34 PM »
Hi everyone, I have went through textbook section 1.5 and found an example quite confusing (pic is attached below). In this example, in order to get the limit (n->∞), we discuss the real part and imaginary part separately. But I don't know how does it solve for the real part. In particular, it says we need to use L'Hopital's rule and we will get x as the result. However, I have been stuck at this computation (the draft in orange color is my computation) and it seems like a dead end. I dont know if I did wrong at the very first step. Besides, when we are using L'Hopital's rule here, are we taking the derivative with respect to n or x? How can we know which is the variable?

22
##### Chapter 2 / LEC 0201 Question
« on: September 18, 2020, 02:20:45 PM »
Hi, we just had a lecture on homogenous ODE this morning, and when I went through the slides afterwards, I have found a confusion in the slide P6(pic is also attached below). I am wondering why the domain is strictly greater than minus root two? Why can't it be smaller than minus root two? Is it because the initial condition of this question is greater than 0 so that our answer needs to be greater than 0? (But why?)

23
##### Chapter 2 / Textbook Section2.1 Example5
« on: September 16, 2020, 09:26:38 PM »
Hi guys, I have been stuck at example 5 of section2.1(page30) for quite a while. In particular, I do not understand why the lower bound of the integral is the initial point t=0. Why can't it be the upper bound?
(example screenshot is attached below)

24
##### Chapter 1 / Question about Inversion
« on: September 16, 2020, 04:11:52 PM »
Hi guys, we talked about inversion during the previous lecture and I am a bit confused by the last slide. So by definition, we have
z-> w=z^-1, and then we can calculate the inversion of any point either by its inverse or its polar form. We have also proved that the inversion of a circle is a vertical straight line in the same slide. But I am a bit confused by the red highlighted part, "inversion is self-inverse". My thought is that an inversion of a circle is a straight line and correspondingly the inversion of that straight line is the original circle. And this property thus makes it self-inverse. I don't know if my understanding is correct so I am writing this post to look for some help. And one more brief question, do we have any restriction on inversion/self-inversion?

(btw slide pic is attached below)

Pages: 1 [2]