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### Topics - Pengyun Li

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##### Quiz 5 / Quiz 5 - 5101 A
« on: November 05, 2020, 07:16:45 PM »
Question: Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $z\cos{\frac{1}{z}}$ at $z = \infty$.

Let $w = \frac{1}{z}$, then we are finding Laurent series for $\frac{\cos{w}}{w}$ at $w = 0$.

$\cos{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n}$,

so $z\cos{\frac{1}{z}} = \frac{\cos{w}}{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n-1} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n-1}$.

Note that $\frac{\cos{w}}{w} = \frac{1}{w}(1-\frac{w^2}{2!}+\frac{w^4}{4!}-...) = \frac{1}{w}- \frac{w}{2!}+\frac{w^3}{4!}-...$,
so the residue of the function is 1 as the coefficient for the term $\frac{1}{w}$ is 1.

$z\cos{\frac{1}{z}} = \frac{1}{\frac{1}{z}}- \frac{\frac{1}{z}}{2!}+\frac{(\frac{1}{z})^3}{4!}-... = z-\frac{1}{z(2!)}+\frac{1}{z^3(4!)}-...$

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##### Quiz 4 / Quiz-5101-D
« on: October 22, 2020, 07:12:31 PM »
Question: Evaluate the given integral using Cauchy’s Formula or Theorem: $\int_{|z|=2}\frac{e^z}{z(z-3)}dz$.

For $z(z-3)=0$, $z=0$ or $z=3$, where only $z=0$ is bounded by $|z|=2$, thus $z_0=0$.

$\int_{|z|=2}\frac{e^z}{z(z-3)}dz = \int_{|z|=2}\frac{\frac{e^z}{z-3}}{z-0}dz$

(By Cauchy's formula)

$= 2\pi i f(z_0)$, where $f(z) = \frac{e^z}{z-3}$ is analytic on $\mathbb{C}$,

Thus, $2\pi i f(z_0) = 2\pi i \cdot \frac{e^0}{0-3} = -\frac{2\pi i}{3}$.

Therefore, $\int_{|z|=2}\frac{e^z}{z(z-3)}dz = -\frac{2\pi i}{3}$.

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##### Quiz 2 / Quiz 2- Lec 5101-A
« on: October 01, 2020, 07:08:28 PM »
$\textbf{Question}$: Find the limit of each function at the given point, or explain why it does not exist: $f(z) = (z-2)log|z-2|$ at $z_0 = 2$.

$\textbf{Answer}$: Since $z_0= 2$, let $z' = z-2$.
$$\lim_{z\to z_0} f(z) = \lim_{z' \to 0} f(z) =z'\log |z'|$$
$$=z' (\ln |z'| +i\cdot 0) = \frac{ln |z'|}{\frac{1}{z'}}$$
By L'Hôpital's Rule, $$= \frac{\frac{1}{z'}}{-\frac{1}{(z')^2}}=\lim_{z'\to 0} (-z') = 0$$.

Therefore, the limit of the function is 0 at $z_0=2$.

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##### Quiz 1 / Quiz 1- Lec 5101-A
« on: September 24, 2020, 07:06:24 PM »
$\textbf{Question:}$ Describe the locus of points z satisfying the given equation: $Re(z^2) = 4$.

$\textbf{Answer:}$

Let $z=x+iy$,

$z^2 = (x+iy) (x+iy) = x^2-y^2+(2xy)i$.

Thus, $Re(z^2) = x^2-y^2=4$.

Therefore, the locus of points z is a hyperbola with equation $x^2-y^2=4$.

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