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### Messages - Ziyi Wang

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##### Quiz 7 / Q7: Wednesday Sitting
« on: March 22, 2020, 02:51:12 PM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the  given annulus: $z^3-3z+1$ in ${1<|z|<2}$
Answer: $f(z)=z^3-3z+1$
On the circle $|z|=1$
$$|z^3+1| \leq |z^3|+|1|=2$$
$$|-3z| = |3z| = 3$$
So $|z^3+1|<|-3z|$. By Roche's Theorem, $f(z)$ and $g(z)=-3z$ have the same number of zeroes with $|z|=1$.
Since $g(z)$ has one zero with $|z|=1$, we know that $f(z)$ has one zero with $|z|=1$.
On the circle $|z|=2$
$$|z^3|=8$$
$$|-3z+1| \leq |3z|+|1| = 7$$
So $|-3z+1|<|z^3|$. By Roche's Theorem, $f(z)$ and $g(z)=z^3$ have the same number of zeroes with $|z|=2$.
Since $g(z)$ has three zeroes with $|z|=1$, we know that $f(z)$ has three zeroes with $|z|=1$.
$$3-1=2$$
Therefore, $f(z)=z^3-3z+1$ has two zeroes in ${1<|z|<2}$.

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##### Quiz 5 / Q5: TUT 0201
« on: March 08, 2020, 06:46:48 PM »
Question: Give the order of each of zeros of the given function.
$$Log(1-z), |z|<1$$
Let $f(z)=Log(1-z), |z|<1$. Let $f(z_0)=0$. Then, we get that $z_0=0$.
Since, $f'(z) = \frac{-1}{1-z}$, we know that $f'(z_0) = -1 \neq 0$.
Therefore, the order of $z_0 = 0$ of the given function is $1$.

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##### Quiz 4 / Q4: TUT 0201
« on: February 17, 2020, 07:12:50 PM »
Question: Find the radius of convergence of the given power series.
$$\sum^\infty _{k=0} \frac{(k!)^2}{(2k)!}(z-2)^k$$
Answer: By ratio test, we have that:
$$\frac{1}{R} = lim_{k\rightarrow \infty} |\frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}|$$
$$= lim_{k\rightarrow \infty} |\frac{\frac{(k+1)^2(k)^2(k-1)^2...}{(2k+2)(2k+1)(2k)...}}{\frac{(k)^2(k-1)^2...}{(2k)(2k-1)...}}|$$
$$= lim_{k\rightarrow \infty} |\frac{(k+1)^2}{(2k+2)(2k+1)}|$$
$$= lim_{k\rightarrow \infty} \frac{k+1}{4k+2}$$
By L’Hospital’s Rule, we know that
$$\frac{1}{R} = lim_{k\rightarrow \infty} \frac{k+1}{4k+2} = lim_{k\rightarrow \infty} \frac{1}{4} = \frac{1}{4}$$
Therefore, we conclude that $R=4$.

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##### Quiz 3 / Q3: TUT 0201
« on: February 07, 2020, 11:46:05 PM »
Question: Directly compute the following line integral:
$$\int _\gamma \frac{dz}{z+4}$$
where $\gamma$ is the circle of radius 1 centered at -4, oriented counterclockwise.
Answer: Let $\gamma (t) = -4 + e^{it}$, where $t \in [0, 2\pi]$.
Then, $\gamma ' (t) = ie ^{it}$.
Then, we compute the line integral:
$$\int _\gamma \frac{dz}{z+4} = \int _0^{2\pi}\frac{ie ^{it} dt}{-4 + e^{it} + 4}$$
$$= \int _0^{2\pi}\frac{ie ^{it}}{ e^{it}}dt$$
$$= \int _0^{2\pi}i dt$$
$$= it \Big|_{t = 0}^{t = 2\pi}$$
$$= 2\pi i$$

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##### Quiz 1 / Q1: TUT 0201
« on: January 29, 2020, 11:07:52 AM »
Questions:
Write the equation of the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$.
Let $z = x + iy$.
Since $z$ is an arbitrary point on the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$, we know that the distance between $z$ and $-1+2i$ is same as the distance between $z$ and $1-2i$.
Then, we can calculate that:
$$|z-(-1+2i)| = |z - (1-2i)|$$
$$|(x + iy)-(-1+2i)| = |(x + iy) - (1-2i)|$$
$$|(x + 1)+(y-2)i| = |(x - 1) + (y + 2)i|$$
$$\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x - 1)^2 + (y + 2)^2}$$
$$(x+1)^2 + (y-2)^2 = (x - 1)^2 + (y + 2)^2$$
$$x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 - 2x + 1 + y^2 + 4y + 4$$
$$4x = 8y$$
$$y = \frac{1}{2} x$$
Then, we write the equation in complex number notation:
$$Re((\frac{1}{2} + i)z) = 0$$

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