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### Messages - Min Gyu Woo

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1
##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5
« on: November 27, 2018, 11:51:00 AM »
Fixed

2
##### 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.

3
##### Term Test 1 / Re: TT1 Problem 4 (night)
« on: October 19, 2018, 11:05:30 AM »
Care to elaborate?
Where did I go wrong?

4
##### 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*}

5
##### Quiz-3 / Re: Q3 TUT 0203
« on: October 12, 2018, 06:17:20 PM »
Our $\gamma (t)$ in this case is

\gamma(t) = 2e^{it}\\

\gamma'(t) = 2ie^{it}

With $0\leq t \leq 2\pi$

Writing the integral in terms of $\gamma$, $\int_{a}^{b} f(\gamma(t))\gamma'(t)dt$

\begin{aligned}

\int_{\gamma}(z^2+3z+4)dz &= \int_{0}^{2\pi}(2e^{2it}+6e^{it}+4)(2ie^{it})dt \\

&= 4i\int_{0}^{2\pi}(e^{3it}+3e^{2it}+4e^{it})dt \\

&= 4i\left[\frac{1}{3i}e^{3it}+\frac{1}{2i}e^{2it}+4\frac{1}{i}e^{it}\right]_0^{2\pi} \\

&= 4\frac{i}{i}\left[\left(\frac{1}{3}e^{6i\pi}+\frac{1}{2}e^{4i\pi}+4e^{2i\pi}\right)-(\frac{1}{3}+\frac{1}{2}+4)\right]\\

&= 0\\

\end{aligned}

Since $e^{i2\pi k} = 1, k\in\mathbb{Z}$

6
##### Thanksgiving bonus / Re: Thanksgiving bonus 4
« on: October 07, 2018, 02:36:47 PM »
Let $f(u,v) = u + iv$ where $u = \cosh{x}\cos{y}$ and $v=\sinh{x}\sin{y}$

Note that

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &= \sinh{x}\cos{y}\\
\frac{\partial v}{\partial y} &= \sinh{x}\cos{y}\\
\frac{\partial u}{\partial y} &= -\cosh{x}\sin{y}\\
\frac{\partial v}{\partial x} &= \cosh{x}\sin{y}

\end{aligned}

\end{equation*}

Since

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &=  \frac{\partial v}{\partial y} \\

\frac{\partial u}{\partial y} &\neq -\frac{\partial v}{\partial x}

\end{aligned}

\end{equation*}

We know that $f$ is not analytic on $D$ by the contrapositve of Cauchy-Riemann Theorem.

So, $f$ is not sourceless or irrotational on $D$.

The error you are talking about is that the textbook used $dx$ instead of $\partial x$

Thus,  the proper equations should be

\begin{equation*}

\begin{aligned}

\int_{\gamma} u dx+v dy &= \iint_{\Omega} \left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right]dxdy=0 \\

\int_{\gamma} u dy - v dx &= \iint_{\Omega} \left[\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right]dxdy=2 \text{ area}(\Omega)

\end{aligned}

\end{equation*}

7
##### Quiz-2 / Re: Q2 TUT 0203
« on: October 05, 2018, 06:15:03 PM »
We know that for a $z\neq0\in\mathbb{C}$, $\log(z)$ is defined to be

\log(z) = \ln|z| + i\arg(z)

Since $1+i\sqrt{3}$ is a complex number

\log(1+i\sqrt{3}) = \ln(2) + i(\frac{\pi}{3}+2\pi{k}), k\in\mathbb{Z}

Which can simplify to

\log(1+i\sqrt{3}) = \log(2) + i(\frac{\pi}{3}+2\pi{k})

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