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Topics - Jiaqi Bi

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Quiz 7 / Lec 0101- Quiz 7 C
« on: December 17, 2020, 06:51:38 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(The graph will be attached.)
(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
\nonumber &=iy+a-cos(y)-isin(y)\\
\nonumber &=(a-cos(y))+i(y-sin(y))
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.

Quiz 5 / Quiz-5-6101 A
« on: November 05, 2020, 08:24:07 AM »
Give the order of each of the zeros of the given function:

Let $w=e^z$, we get $w^2-3w-4=(w-4)(w+1)$
Then resubstitute $e^z=w$
We get $(e^z-4)(e^z+1)=0 \Rightarrow e^z=4\ \text{and}\ e^z=-1$
Solve for $z$, we get $z=log(4)=ln(4)+i(2k\pi)\ \text{and}\ z=log(-1)
=\pi i+i(2k\pi)
To find orders: $f'(z)=2e^{2z}-3e^z$
Substitute $e^z=4\ \text{and}\ -1$ into $f'(z)$
$2\times 16-3\times 4 = 20 \neq 0, 2+3=5\neq 0$
Since first derivative does not equal to $0$ for both $z$, we conclude order $= 1$
Hence, the order is $1$ for $z=ln(4)+i(2k\pi)\ \text{and}\ z=i(\pi+2k\pi)$

Quiz 4 / Quiz-4-6101 C
« on: October 23, 2020, 08:37:43 AM »
Question: $$\int_\gamma sin(z)dz,$$
where $\gamma$ is any curve joining $i$ to $\pi$
&\int_\gamma sin(z)dz\\
&=\int_{i}^{\pi} sin(z)dz\\
&=-cos(z) \Big|_{i}^{\pi}\\

Quiz 2 / Quiz 2 Session 6101
« on: October 01, 2020, 04:46:51 PM »
Question: $$\sum_{n=1}^{\infty} \frac{1}{n} (\frac{1+i}{\sqrt{2}})^{n}$$
\lim_{n \to \infty} \frac{1}{n}(\frac{1+i}{\sqrt{2}})^{n} \\
=\lim_{n \to \infty} \frac{1}{n}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
=\lim_{n \to \infty} \frac{1}{n}\cdot \lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
Since $$\lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\
=\lim_{n \to \infty}(\frac{\sqrt{2}\ (cos\frac{\pi}{4}+isin\frac{\pi}{4})}{\sqrt{2}})^{n}\\
=\lim_{n \to \infty}(cos \frac{n\pi}{4}+isin \frac{n\pi}{4}) \ \ (\text{By De Moivres Theorem})\\
=\text{DNE}\ \text{and it diverges}
Hence, the series diverges.

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