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### Messages - Min Gyu Woo

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1
##### MAT334--Lectures & Home Assignments / Sample Final Solutions
« on: December 11, 2018, 08:38:32 PM »
Could the prof go over the solutions for the sample final? I think some of them are wrong...

2
##### MAT334--Lectures & Home Assignments / Re: FE Sample Question 4 (a)
« on: December 08, 2018, 02:14:56 PM »
Can you approach this method another way without the hint?

I.e) using the three fixed points (choose the last one to be any point that matches the conditions)

3
##### MAT334--Lectures & Home Assignments / Re: Final Exam Scope?
« on: December 06, 2018, 03:38:21 PM »
So we need to know how to do TT1 + TT2 + Q7 ***AND*** the Sample Final correct?

Also, the Sample Final contains questions about the stretch and rotation angle of a mobius transformation. Isn't that part of Chapter 3.4? I don't see anything in 3.3 that discusses rotation and stretch.

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##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5
« on: November 27, 2018, 11:51:00 AM »
Fixed

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##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5
« on: November 27, 2018, 10:54:10 AM »
Let $f(z) = e^2z$ and $g(z) = e^z$.

We have

$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$

So, $f(z) = 0$ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)

$$f(z) = e^2z=0$$
$$z = 0$$

Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.

Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$

Then,

Call $h(z) = f(z) - g(z) = e^x - e^2 x$

Note that:

$$h(0) = 1$$

$$h(1) = e - e^2 <0$$

By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.

I.e. there is a REAL ROOT x where $0<x<1$.

We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.

Thus, within $|z| <1$ there is only one root, and that root is real.

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##### MAT334--Lectures & Home Assignments / Re: 2.6 Q22
« on: November 22, 2018, 02:32:44 PM »
The steps of getting to that final equation. Not the formula.

7
##### MAT334--Lectures & Home Assignments / Do we have to know Deformation Theorem for the test?
« on: November 22, 2018, 11:42:45 AM »
It's not in the textbook but was covered in lecture. Also, most of those questions can be solved using other tactics correct?

8
##### MAT334--Lectures & Home Assignments / Re: 2.5 - Q23
« on: November 20, 2018, 10:05:16 PM »
Is it just me or is the entirety of Q23 answer key wrong...

9
##### MAT334--Lectures & Home Assignments / Re: 2.5 - Q23
« on: November 20, 2018, 07:02:57 PM »
Can you guys explain why you can use (12) for z = -1, and (13) for z=2?

Following the definition in the textbook you can't use either (12) or (13) because |z_j| < r and |z_j| > R aren't satisfied.

10
##### MAT334--Misc / Re: Cheating during the quiz
« on: November 02, 2018, 11:14:32 AM »
Oh sorry, maybe I'll pull out my phone and record them during the quizzes or something.

On a serious note, IN MY OPINION that's what it looks like.

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##### MAT334--Misc / Cheating during the quiz
« on: November 02, 2018, 12:38:21 AM »
Hello Professor,

Could you tell the TAs to actually punish people who cheat? Some TAs seem to be either oblivious or ignoring the fact that a large portion of the class discuss answers.

12
##### Term Test 1 / Re: TT1 Problem 4 (night)
« on: October 19, 2018, 11:05:30 AM »
Care to elaborate?
Where did I go wrong?

13
##### Term Test 1 / Re: TT1 Problem 4 (night)
« on: October 19, 2018, 08:55:00 AM »
The equation for the curve is defined to be

$$L(t) = 3i + 3e^{it}$$ for $\frac{\pi}{2} \geq t\geq \frac{-\pi}{2}$

Also, $L'(t)=3ie^{it}$

We're given that $f(z)=\overline{z}^2$ so all we have to do is parametrize the equation:

\begin{align*}

\int_{L}f(z)dz&=\int_{\pi/2}^{-\pi/2}(\overline{3i+3e^{it}})^2(3ie^{it})dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(i-e^{-it})^2(e^{it}) dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-1-2ie^{-it}+e^{-2it})(e^{it})dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-e^{it}-2i+e^{-it})dt \\
&=-3^3i\int_{\pi/2}^{-\pi/2}(e^{it}-e^{-it}+2i)dt \\
&=-3^3i\left[\frac{1}{i}e^{it}+\frac{1}{i}e^{-it}+2it\right]^{-\pi/2}_{\pi/2} \\
&=\frac{-3^3i}{i}\left[e^{it}+e^{-it}+2i^2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3\left[e^{it}+e^{-it}-2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3[(e^{-i\pi/2}+e^{i\pi/2}-2(-\pi/2))-(e^{i\pi/2}+e^{-i\pi/2}-2(\pi/2))] \\
&= -3^3[(-i+i+\pi)-(i-i-\pi)] \\
&= -3^3[\pi+\pi] \\
&= -54\pi
\end{align*}

14
##### MAT334--Misc / Changing the Scope of the test
« on: October 17, 2018, 04:50:05 PM »
Professor,

Don't you think it's a little bit late to change the coverage of the test to include 2.2 all of a sudden?

15
##### MAT334--Lectures & Home Assignments / Re: Chapter 1.6 PG 63 ex Example 8
« on: October 17, 2018, 01:13:09 PM »
Triangle Inequality is

$$|z+w| \leq |z| + |w|$$

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