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### Messages - Kuba Wernerowski

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##### Chapter 3 / Child Boards for Quizzes 6/7 and Tests 3/4
« on: November 20, 2020, 10:19:26 AM »
The child boards for the last two quizzes and last two terms tests aren't available as of this post; could these be made available to us soon?

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##### Quiz 4 / Quiz-5101-A
« on: October 23, 2020, 11:17:53 AM »
Evaluate the given integral using the technique of Example 10 of Section 2.3: $$\int_\gamma \frac{dz}{z^2},$$ where $\gamma$ is any curve in $\{ z: Re \, z \geq 0, z \neq 0 \},$ joining $-i$ to $1+i$.

Solution:

$F(z) = \frac{-1}{z}$, where $F'(z) = f(z) = \frac{1}{z^2}$.

Note that $F$ is analytic whenever $z \neq 0$. Therefore, $F$ is analytic on $\gamma$.

So we have $$\int_\gamma f(z) dz = \int_\gamma F'(z) dz$$
\begin{align*}
\int_\gamma F'(z) dz &= F(\text{end point}) - F(\text{initial point}) \\
&= F(1 + i) - F(-i) \\
&= \frac{-1}{1+i} - \left(\frac{-1}{i}\right) \\
&= \frac{-i + 1 + i}{(1+i)i} \\
&= \frac{1}{i-1} \\
&= \frac{1}{i-1} \frac{i+1}{i+1} \\
&= -\frac{1 + i}{2} \\

\end{align*}

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##### Chapter 1 / Re: Section 1.4: Question 29 Proof Check
« on: October 07, 2020, 09:38:32 AM »
For any polygonal segment $P_i P_j$ where that's the case, could we divide that segment into smaller sub-segments $P_{i_1} P_{i_2} \cdots P_{i_{n-1}}P_j$ and put open, overlapping discs over each sub-segments end/start points?

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##### Chapter 1 / Section 1.4: Question 29 Proof Check
« on: October 02, 2020, 01:35:40 PM »
My proof follows roughly the same logic as the outline in the textbook, but I'm not sure if it's quite rigorous enough.

Choose $p_a, p_b \in D$. Since $D$ is a domain, $\exists$ a polygonal curve $P_1 P_2 \cup P_2 P_3 \cup \cdots P_{n-1}P_n$ connecting $p_a$ and $p_b$.

Since $D$ is open, each point line segment $P_1, \ldots P_n$ have an open disc $A_j$ centered at $A_j$ where $j = 1, 2, \ldots, n$.

Construct the polygonal curve such that for each pair of endpoints, $P_j, P_{j+1}$, their respective open discs $A_j, A_{j+1}$ have the property that $A_j \cap A_{j+1} ≠ \emptyset$ for $j=1, 2,\ldots, n-1$.

Then, since $u$ has the property that for each of those open discs, $u(A_j)$ = some constant, $c_j$.

$u(A_1) = c_1$, $u(A_2) = c_2$, and given that $A_1 \cap A_2 \neq \emptyset$, then $c_1 = c_2$.

Repeat this argument for each pair of $A_j$ and $A_{j+1}$ until we reach $A_{n-1} \cap A_n \neq \emptyset \implies u(A_{n-1}) = u(A_n)$.

To conclude the proof, $p_a \in A_1$ and $p_b \in A_n$ meaning that $u(p_a) = u(p_b)$, and since $p_a, p_b$ were chosen arbitrarily, $u$ is constant on $D.$.

Any feedback / criticisms are much appreciated

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##### Quiz 2 / LEC0101 - Quiz 2 D
« on: October 02, 2020, 11:16:10 AM »
$\textbf{Problem}$ (3pt). Find all the value(s) of the given expression $$i^{\sqrt{3}}.$$
\begin{align*}\
i^{\sqrt{3}} &= e^{\ln(i)^{\sqrt{3}}} \\
&= e^{\sqrt{3} \ln(i)} \\
&= e^{\sqrt{3}(\ln \lvert{i}\rvert + i \arg{i})} \qquad \quad \ln(i) = \ln \lvert i \rvert + i \arg{i} \text{ since } i \in \mathbb{C}.\\
&= e^{\sqrt{3}(\ln{1} \, + \, i \left(\frac{\pi}{2} + 2\pi k \right))} \qquad \, k \in \mathbb{Z} \\
&= e^{\sqrt{3}(i \left(\frac{\pi}{2} + 2\pi k \right))} \\
&= \cos{\sqrt{3} \left(\frac{\pi}{2} + 2 \pi k \right)} + i \sin{\sqrt{3} \left(\frac{\pi}{2} + 2 \pi k \right)}
\end{align*}

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##### Chapter 1 / Re: why complex plane closed
« on: October 01, 2020, 09:05:13 PM »
I used to be really confused by this until I really looked at the definitions of closed and open sets.

By definition, a set $S$ is closed if $S^c$ is open, and vice-versa.

The empty set $\emptyset$ trivially has nothing but interior points, and also trivially includes its boundary. This took me a while to internalize, but I think it's easiest to just ¯\_(ツ)_/¯  and accept it.

Since $\emptyset$ is open, then $\emptyset^c = \mathbb{C}$ is closed. And since $\emptyset$ is closed, then $\emptyset^c = \mathbb{C}$ is open.

Therefore, $\mathbb{C}$ is both open and closed.

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##### Quiz 1 / LEC0101 - Quiz 1 C
« on: September 25, 2020, 12:37:22 PM »
$\textit{Find all solutions of the given equation:}$ $$(z+1)^4 = 1 - i.$$

$1 - i$ on the xy plane is equivalent to $(1, -1)$, which has an angle $\theta = \frac{-\pi}{4}$ and length $\sqrt{2} = 2^{1/2}$.

In polar representation, $$1-i = 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \quad k \in \mathbb{Z}.$$

Back to the original question,

\begin{align*}

(z+1)^4 &= 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \\
z+1 &= (2^{1/2} e^{i(- \pi/4 + 2k \pi)})^{1/4} \\
z &= 2^{1/8} e^{i(- \frac{\pi}{16} + \frac{1}{2}k \pi)} - 1
\end{align*}

We only need to solve for $k = 0, 1, 2, 3$ since $(z+1)^4$ has 4 roots.

$$z = \begin{cases} 2^{1/8} e^{i \left(- \frac{\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{-\pi}{16} - i \sin\frac{-\pi}{16}) & \text{for } k=0, \\ 2^{1/8} e^{i \left(\frac{7\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{7\pi}{16} - i \sin\frac{7\pi}{16}) & \text{for } k=1, \\ 2^{1/8} e^{i \left(\frac{15\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{15\pi}{16} - i \sin\frac{15\pi}{16}) & \text{for } k=2, \\ 2^{1/8} e^{i \left(\frac{23\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{23\pi}{16} - i \sin\frac{23\pi}{16}) & \text{for } k=3 \end{cases}$$

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