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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...
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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)
I.e) using the three fixed points (choose the last one to be any point that matches the conditions)
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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.
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.
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.
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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?
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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...
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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.
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.
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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.
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.
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.
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Term Test 1 / Re: TT1 Problem 4 (night)
« on: October 19, 2018, 11:05:30 AM »
Care to elaborate?
Where did I go wrong?
Where did I go wrong?
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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 \\
&=3i\int_{\pi/2}^{-\pi/2}(-3i+e^{-it})^2(e^{it})dt \qquad\color{red}{\text{Misprint}}\\
&=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*}
$$L(t) = 3i + 3e^{it}$$ 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 \\
&=3i\int_{\pi/2}^{-\pi/2}(-3i+e^{-it})^2(e^{it})dt \qquad\color{red}{\text{Misprint}}\\
&=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*}
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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?
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?
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MAT334--Lectures & Home Assignments / Re: Chapter 1.6 PG 63 ex Example 8
« on: October 17, 2018, 01:13:09 PM »
Triangle Inequality is
$$|z+w| \leq |z| + |w|$$
$$|z+w| \leq |z| + |w|$$
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