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Topics - Ende Jin

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MAT334--Lectures & Home Assignments / Residue for FE P3
« on: December 12, 2018, 03:27:33 PM »
I don't think that is correct because if let $\sin^2(z) = (z-n\pi)^2h(z)$ where $h(n\pi) \neq 0$, then $\frac{1}{h(z)}$ is analytic at neighbourhood of $n\pi$.
Thus $Res(\frac{\cos(\frac{z}{6})}{\sin^2(z)}, n\pi) = Res(\frac{\cos(\frac{z}{6})}{(z-n\pi)^2h(z)}, n\pi) = $ coefficient of $(z-n\pi)^{1}$ for function $ \frac{\cos(\frac{z}{6})}{h(z)}$
while $cos(\frac{z}{6}) =a_0 + a_1(z-n\pi)^{1} + O((z-n\pi)^2) $ and $\frac{1}{h(z)} = b_0 + b_1(z-n\pi)^{1} + O((z-n\pi)^2) $
thus the coefficient for the fraction at $(z-n\pi)^{1}$ is $a_0b_1 + a_1b_0$, which I get is $ 0 + (- \frac{1}{6} sin(\frac{n\pi}{6})\frac{1}{\cos^2(n\pi)})$

Just asking for a check of this idea.

MAT334--Lectures & Home Assignments / Sample Final, Q4, (c)
« on: December 06, 2018, 12:15:20 PM »
I just want to ask, if that is the answer. Because the author just took off the absolute value without reason ($z \in \mathbb{C}$ right?) and when dealing with $arg(f'(z))$ I don't understand why the simplification is necessary. Why not just write $arg(f'(z)) = arg(\frac{-24}{(1+5z)^2})$ Why it has to move $\pi$ out?

MAT334--Lectures & Home Assignments / 3.3 Q7, b,c,d
« on: December 05, 2018, 08:28:32 PM »
I have tried (b) and had some clue about it. Basically I choose the intersection of the original to map to the intersection in the codomain. That means choosing $\{0, \infty\} \mapsto \{0, 1\}$. After that, I can get only a form like $z \mapsto \frac{az}{az+d}$. And then I give it $x \in \mathbb{R}$ and $iy, y \in \mathbb{R}$ to ensure the two condition is satisfied. With that, I can get $ad \in \mathbb{R},\bar{a}d \in \mathbb{R}$. And I arbitrarily chose  a solution and it works.

My question is if there is a general way to do this kind of question? Because in the above, there are too many guessing.

MAT334--Lectures & Home Assignments / Final Exam Scope?
« on: November 29, 2018, 07:53:08 PM »
What will be covered in the final? Since on the schedule, a conformal mapping is still something marked as "(if permits)".

MAT334--Lectures & Home Assignments / 3.2 Q1?
« on: November 27, 2018, 12:05:43 PM »
There is no example in the book.
The only way I know is to translate it into multivariable calculus and use Lagrange multiplier. However, is there an easier way?

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.

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?

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.

I found that the definition of "arg" and "Arg" in the book is different from that introduced in the lecture (exactly opposite) (on page 7).
I remember in the lecture, the "arg" is the one always lies in $(-\pi, \pi]$
Which one should I use?

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