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### Messages - Min Gyu Woo

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16
##### MAT334--Lectures & Home Assignments / Re: The Partial Derivative of f = u + iv on textbook
« on: October 16, 2018, 10:29:54 PM »
It seems like the equation in 2.1 is

$$\frac{df(z)}{dz}$$

while the equation in 1.6 is

$$\frac{\partial f(p,q)}{\partial x} \text{ and } \frac{\partial f(p,q)}{\partial y}$$

Basically, the 2.1 equation talks about the derivative while 1.6 equations talk about the two separate partial derivatives.

17
##### MAT334--Lectures & Home Assignments / Chapter 1.6 PG 63 ex Example 8
« on: October 16, 2018, 10:17:54 PM »
Can someone explain how to use the triangle inequality to end up with

$$|z^2+4|\geq |z|^2-4$$

18
##### MAT334--Lectures & Home Assignments / Chapter 2.1 PG 80
« on: October 16, 2018, 08:32:30 PM »
There's this little portion of text before the defining Laplace's Equation that I am confused about.

"A word of warning is merited here: Not every function $u(x,y)$ is the real part of an analytic function".

If we always define $f = u +iv$, doesn't this guarentee that $u(x,y)=\text{Re}(f)$?

19
##### Quiz-3 / Re: Q3 TUT 0203
« on: October 12, 2018, 06:17:20 PM »
Our $\gamma (t)$ in this case is

\begin{equation}

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

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

\end{equation}

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

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

\begin{equation}

\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}

\end{equation}

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

20
##### 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*}

21
##### 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

\begin{equation}

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

\end{equation}

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

\begin{equation}

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

\end{equation}

Which can simplify to

\begin{equation}

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

\end{equation}

22
##### MAT334--Lectures & Home Assignments / Re: Section 1.4 Question 1
« on: October 02, 2018, 02:27:13 PM »
"This sequence converges to 0 because the absolute value converges to 0."

Where did you get that information?

The textbook states on pg 34 that,

If $z_n \rightarrow A$ then $|z_n| \rightarrow |A|$.

Does the converse work as well?

23
##### MAT334--Lectures & Home Assignments / Re: Hint for Question 20 in Section 1.2
« on: September 27, 2018, 09:17:48 PM »
I thought of it this way. Hope this helps.

24
##### MAT334--Lectures & Home Assignments / Re: Hint for Question 20 in Section 1.2
« on: September 24, 2018, 03:39:43 PM »
So if we set the origin to be $z_1$,

Is it enough to show that the $tz_1 + (1-t)z_2$ follows the equation of the line

$Re(-(t-1)(1+i)z_2 + 0)=0$ which simplifies to $y=x$ ?  How that?! V.I.

EDIT:

$Re(-(t-1)(1+i)z_2) =0$

If $a=(1-t)(1+i)=(1-t)+i(1-t) \in \mathbb{C}$

We can simplify to

$Re(az_2 + 0) = (1-t)x -(1-t)y + Re(0) = 0$

Then we get

$(1-t)x=(1-t)y$

$x=y$

NOTE: When $t=1$ we get $z_1 = 0$ which we already have if we centered $z_1$ at the origin.

25
##### MAT334--Lectures & Home Assignments / Re: Section 1.2 Question 18
« on: September 24, 2018, 02:15:00 PM »
Isn't $Re(a\overline{c}) = AC + BD$?

26
##### MAT334--Lectures & Home Assignments / Re: Section 1.4 Example 8
« on: September 21, 2018, 07:55:57 AM »
Could you explain more about the trig function vanishing?

27
##### MAT334--Lectures & Home Assignments / Section 1.4 Example 8
« on: September 20, 2018, 08:15:41 PM »
Can someone prove this using the definition of limits for sequences:

$\lim_{n\rightarrow\infty} (1/n)(\cos{(n\pi/4)}+i\sin{(n\pi/4)}) = 0$

28
##### MAT334--Lectures & Home Assignments / Textbook error?
« on: September 14, 2018, 02:41:21 PM »
At the bottom of the page 19, it states that

Re$[(\overline{z - i\alpha})(z-s)]=0$

to show that the two segments are perpendicular.

However, on the next page it continues the proof by writing

$2$Re$[(\overline{z - i\alpha})(z-s)]=0$

Where did the $2$ come from?

29
##### MAT334--Lectures & Home Assignments / First example in class
« on: September 14, 2018, 11:28:19 AM »
For the example with $z^3 - 3z$

Where did the $M$ come from?

Also what limit were we computing with the function $f(z) = z^3 - 3z$?

Also why is $|z^2 +zz^{*}+z^{*2}|$ less than or equal to $3M^2$ ?

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