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### Messages - Muyao Chen

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1
##### Final Exam / Re: FE-P1
« on: December 18, 2018, 01:46:33 PM »
$$f(z) = \frac{A}{z-(-1+i)} + \frac{B}{z-(-1-i)}$$

Solve A, B
$$A = -\frac{i}{2}, B = \frac{i}{2}$$

Then
$$f(z) = -\frac{i}{2}\frac{1}{z-(-1+i)} +\frac{i}{2} \frac{1}{z-(-1-i)}$$

When $\mid z \mid = r$

$$f(z) = -\frac{i}{2} \frac{1}{-1+i} \frac{1}{\frac{z}{-1+i}-1} + \frac{i}{2}\frac{1}{-1-i}\frac{1}{\frac{z}{-1-i}-1}$$
$$= \frac{i}{2} \frac{1}{-1+i} \frac{1}{1 - \frac{z}{-1+i}} - \frac{i}{2}\frac{1}{-1-i}\frac{1}{1 - \frac{z}{-1-i}}$$
$$= \frac{i}{2} \frac{1}{-1+i} \sum_{n=0}^{\infty} (\frac{z}{-1+i})^{n} - \frac{i}{2}\frac{1}{-1-i}\sum_{n=0}^{\infty}(\frac{z}{-1-i})^{n}$$

So that converge at
$$\mid \frac{z}{-1+i} \mid < 1$$
$$\mid z \mid < \sqrt{2}$$

So that not converge at $\mid z \mid = \sqrt{2}$

When $\mid z \mid = R$
$$f(z) = -\frac{i}{2} \frac{1}{z} \frac{1}{ 1 - \frac{-1+i}{z}} + \frac{i}{2}\frac{1}{z}\frac{1}{1 - \frac{-1-i}{z}}$$
$$= \frac{i}{2} \frac{1}{z} \sum_{n=0}^{\infty} (\frac{-1+i}{z})^{n} + \frac{i}{2}\frac{1}{z}\sum_{n=0}^{\infty}(\frac{-1-i}{z})^{n}$$

So that converge at
$$\mid \frac{-1+i}{z} \mid < 1$$
$$\mid z \mid > \sqrt{2}$$

So that not converge at $\mid z \mid = \sqrt{2}$

2
##### Quiz-7 / Re: Q7 TUT 0102
« on: December 01, 2018, 12:25:02 PM »
Because as R $\rightarrow \infty$, others goes to 0.

3
##### Quiz-7 / Re: Q7 TUT 5201
« on: November 30, 2018, 11:32:41 PM »
$f(z) = z^{4} -3z^{2} + 3$ substitute $w = z^{2}$
$$z^{4} -3z^{2} + 3 = w^{2} - 3w + 3 = 0$$

$w = \frac{3 \pm i \sqrt 3}{2}$, so $z=\sqrt{\frac{3 \pm i \sqrt 3}{2}}$

For z in $[0, R]$:
$$f(x)= x^{4}-3x^{2}+ 3$$
$f(0) = 3$, then $arg(f(z)) = 0$

For $z = Re^{it}$ And $0 \leq t \leq \frac{\pi}{2}$:
$$f(Re^{it}) = R^{4}e^{4it} - eRe^{2it} + 3 = R^{4}(e^{4it} - \frac{3e^{2it}}{R^{3}} + \frac{3}{R^{4}}) = R^{4}e^{4it}(1 - \frac{3}{R^{3} e^{2it}} +\frac{3}{R^{4}})$$
As $R \rightarrow \infty$
$f(z)= R^{4}e^{4it}$ along 2, so that t goes from 0 to $\frac{\pi}{2}$, then $arg(f(z))$ goes from $4 *0 = 0$ to $4* \frac{\pi}{2}$ = 2$\pi$

For $z = iy$, And $0 \leq y \leq R$:
$$f(iy) = y^{4} - 3y + 3$$
$$f(0) = 3$$
Then $arg(f(z)) = 0$
Then $$\triangle arg(f(z)) = 0 +2 \pi + 0 = 2 \pi$$
$$\frac{1}{2 \pi}[\triangle arg(f(z))] = N_{0} - N_{p} = N_{0} = \frac{1}{2 \pi}2 \pi = 1$$

Then only one solution of the function in the first quadrant.

4
##### Quiz-7 / Re: Q7 TUT 0102
« on: November 30, 2018, 10:07:26 PM »
$$f(z) = 2z^{4} - 2iz^{3} + z^{2}+ 2iz -1$$

Consider contour in the upper half-plane with radius R.
let $$z = Re^{i \theta}$$
with$$\theta \in [0, \pi]$$
Then$$f(Re^{i \theta}) = R^{4}( (2e^{4i \theta}) + O( \frac{1}{R} ))$$
Then$$argf(Re^{i \theta}) = 4 \pi$$
On the real axis,$$f(x) = 2x^{4}+x^{2}-1-2ix(x^{2}-1)$$
Then zeros for the real part is:$$x = \frac{-1 \pm 3}{4}$$
for the imaginary part is:$$x = \pm 1, 0$$
When $$x < -1$$it's in first quadratic.
When $$-1 < x < \frac{-1 \pm 3}{4}$$it's moved to fourth quadrant.
So that$$\triangle argf(x) = -2 \pi$$

Then$$\frac{1}{2 \pi} (\triangle argf(z)) = 1$$
So that calculate the number of zeroes of the following function in the upper half-plane is 1.

5
##### Quiz-7 / Re: Q7 TUT 0201
« on: November 30, 2018, 09:52:48 PM »
$$p(z) = ze^{z} - \frac{1}{4}$$
Since $$f(0) \neq 0$$
It would be same as finding the number of zeros in
$$\mid z \mid < 2$$
On
$$\mid z \mid = 2$$
$$\mid ze^{z}\mid = 2e^{Re(z)} > 2e^{-2} = 0.276 > \frac{1}{4}$$
So p(z) and $ze^{z}$ have the same number of zero in $\mid z \mid < 2$.
So that number of zeros of f(z)is one in $0 < \mid z \mid < 2$.

6
##### Term Test 2 / Re: TT2B Problem 5
« on: November 24, 2018, 11:23:38 AM »
$$f(z) = \frac{1}{z-3} - \frac{1}{z-5}$$

(a) $\mid z \mid$ $<$ 3

$$f(z) = \frac{1}{3} - \frac{1}{1 - \frac{z}{3}} - \frac{1}{5} \frac{1}{1+ \frac{z}{5}} = - \frac{1}{3} \sum_{n=0}^{\infty} \frac {z}{3}^{n} - \frac{1}{5} \sum_{n=0}^{\infty} - \frac {z}{5}^{n} = \sum_{n=0}^{\infty} (- \frac{1}{3^{n +1}} - \frac{(-1)^{n}}{5^{n+1}}) z^{n}$$

(b) 3 $<$ $\mid z \mid$ $<$ 5

$$f(z) = \frac{1}{z} \frac{1}{1 - \frac{3}{z}} - \frac{1}{5} \frac{1}{1+ \frac{z}{5}} = \frac{1}{z} \sum_{n=0}^{\infty} \frac {3}{z}^{n} - \frac{1}{5} \sum_{n=0}^{\infty} - \frac {z}{5}^{n} = \sum_{n=0}^{\infty} \frac{3^{n}}{z^{n+1}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n+1}}z^{n} = \sum_{n = - \infty}^{1} 3^{-n} z^{n-1} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{5^{n+1}}z^{n}$$

(c) $\mid z \mid$ $>$ 5

$$f(z) = \frac{1}{z} \frac{1}{1 - \frac{3}{z}} - \frac{1}{z} \frac{1}{1+ \frac{5}{z}} = \frac{1}{z} \sum_{n=0}^{\infty} \frac {3}{z}^{n} - \frac{1}{z} \sum_{n=0}^{\infty} - \frac {5^{n}}{z^{n}} = \sum_{n=0}^{\infty} (\frac{3^{n}}{z^{n+1}} - (-1) ^{n} \frac{5^{n}}{z^{n+1}}) = \sum_{n = - \infty}^{0} (3^{-n} - (-1)^{-n} 5^{-n} ) z^{n-1}$$

7
##### Quiz-6 / Re: Q6 TUT 5201
« on: November 17, 2018, 08:58:45 PM »
The first four terms of the Laurent series is:

$\frac{2}{z^{2}}$ + $\frac{1}{6}$ +$\frac{z^{2}}{120}$ +$\frac{z^{4}}{3024}$ +...

8
##### Quiz-6 / Re: Q6 TUT 0203
« on: November 17, 2018, 08:50:04 PM »
Write

f(z) =  $\frac{g(z)}{(z- z_{0})^{l}}$

so

f '(z) = $\frac{g'(z)(z- z_{0})^{l} - g(z)(z- z_{0})^{l -1}}{(z- z_{0})^2}$

=g'(z) $\frac{1}{(z- z_{0})^{l}}$ - g(z)l$\frac{1}{(z- z_{0})^{l + 1}}$

so that

$\frac{f'}{f}$ = $\frac{g'(z) \frac{1}{(z- z_{0})^{l}} - g(z)l\frac{1}{(z- z_{0})^{l + 1}}}{\frac{g(z)}{(z- z_{0})^{l}}}$

= $\frac{g'}{g}$ - $\frac{l}{z-z_{0}}$

Then

$Res(\frac{f'}{f}, z_{0})$ = $Res(\frac{g'}{g}$ - $\frac{l}{z-z_{0}}, z_{0})$ =  $Res(\frac{- l}{z-z_{0}}, z_{0})$  = $\frac{- l}{(z-z_{0})'}$ = $\frac{-l}{1}$ = $-l$

9
##### Quiz-6 / Re: Q6 TUT 0301
« on: November 17, 2018, 06:07:07 PM »
$\frac{sin z}{(z-π)^2}$ = $\frac{-sin (z- \pi)}{(z-π)^2}$ = $-(z - \pi)^{-2}sin (z- \pi)$
= $-(z - \pi)^{-2} \sum_n^∞ \frac{(-1)^{n}(z- \pi)^{2n+1}}{(2n+1)!}$
= $\sum_n^∞ \frac{(-1)^{n+1}(z- \pi)^{2n-1}}{(2n+1)!}$
= $\frac{-1}{z - \pi} +\sum_k^∞ (-1)^{k+1} \frac{(z- \pi)^{2k+1}}{(2n+1)!}$
$Res(f, \pi) = -1$

10
##### Quiz-6 / Re: Q6 TUT 0102
« on: November 17, 2018, 05:58:19 PM »
$\frac{z}{(sinz)^2}$
sinz = 0, Z = 0
numerator f(z)= z, f(0) = 0,  f'(z) $\neq$ 0, so order = 1
g(z)= $(sinz)^2$, g(0)= 0, g'(z) = 2sinzcosz = sin2z,
g''(z) = 2cos2z $\neq$ 0, so order = 2.
so it's simple pole
Then
$\frac{z}{(sinz)^2}$ = $a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...$
$\frac{z}{(Z - \frac{Z^{3}}{3!} +\frac{Z^{5}}{5!} -...)^2}$ = $a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...$
$Z = (Z - \frac{Z^{3}}{3!} +\frac{Z^{5}}{5!} -...)^2(a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...)$
Then
$\frac{z}{(sinz)^2}$ =  $\frac{1}{z}$ + $\frac{z}{3}$ + $\frac{z^{3}}{15}$ + $\frac{2Z^{5}}{189}$ +...
$Res(f;0)= 1$

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