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### Messages - Jiayue Wu

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##### Quiz 5 / Quiz 5 TT 0401
« on: March 03, 2020, 05:04:55 PM »
Find the first four terms in power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
$$[Log(1-z)]^2 \ about \ z_0=0$$

$$[Log(1-z)]' = -\frac{1}{1-z} = -\sum_{n=0}^{\infty}z^n, \text{ valid at } |z| <1$$
$$Log(1-z) =- \int\sum_{n=0}^{\infty}z^n dz =-\sum_{n=0}^{\infty}\int z^n dz = -\sum_{n=0}^{\infty}\frac{z^n+1}{n+1} =- \sum_{m=1}^{\infty}\frac{z^m}{m}$$
$$[Log(1-z)]^2 = [- \sum_{m=1}^{\infty}\frac{z^m}{m}]^2 = [\sum_{m=1}^{\infty}\frac{z^m}{m}]^2 \\=(z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+...)(z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+...) \\=z^2+z^3+\frac{11}{12}z^4+\frac{5}{6}z^5 +...$$

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##### Quiz 3 / Quiz 3 TUT 0401
« on: February 12, 2020, 05:54:31 PM »
Question: Evaluate the given integral using Cauchy's Formula or Theorem.
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz$$
We can find that on the region $|z| = 2$, $F(z) = \frac{e^z}{z(z-3)}$ not continuous at z = 0. Therefore I'll apply Cauchy's Formula.
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz = \int_{|z| = 2} \frac{e^z /(z-3)}{z}dz$$
$$\implies f(z) = \frac{e^z}{z-3}, f(0) = -\frac{1}{3}$$
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz = 2\pi i f(0) = -\frac{2\pi i}{3}$$

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##### Quiz 2 / Re: TUT0301 Quiz2
« on: February 12, 2020, 05:39:51 PM »
$\lim \limits_{z_{0} \to i}\ f(x) = \lim \limits_{z_{0} \to i}\ |1-z|^2 = |1-i|^2 = 1^2+-(1)^2 = 2$

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##### Quiz 1 / Quiz1 Tut0401
« on: February 12, 2020, 05:29:12 PM »
Question: Describe the locus of points z satisfying the given equation
$$|z-1|^2 = |z+1|^2 +6$$
$$z=x+iy$$
$$z-1 = (x-1)+iy$$
$$z+1 = (x+1)+iy$$
$$|z-1|^2 = |z+1|^2+6 \implies (x-1)^2+y^2 = (x+1)^2 + y^2 +6$$
$$-4x=6\implies x =- \frac{3}{2}\\ z=- \frac{3}{2}+iy, y\in \mathbb{R}$$
The locus of points z: z is a verticle line with $x =- \frac{3}{2}$

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##### Quiz-2 / MAT 244 TUT 0201 Quiz 2
« on: October 04, 2019, 06:25:19 PM »
show that the given equation is not exact but becomes exact
when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3+x(1+y^2)y'=0, u(x.y)=\frac{1}{xy^3}$$

Solution:
This is not exact because
$$\frac{\partial}{\partial y}(x^2y^3)=3x^2y^2 \neq \frac{\partial}{\partial x}[x(1+y^2)]=q=y^2$$
However, multiplyin by $\mu$, we get:     $$x+\frac{1+y^2}{y^3}y'=0$$
it is exact since $$\frac{\partial}{\partial y}(x) = \frac{\partial}{\partial x}(\frac{1+y^2}{y^3})=0$$
Next, I will find the function $\phi$.
$$\frac{\partial \phi}{\partial x}= x$$
$$\phi = \int x dx = \frac{x^2}{2}+h(y)$$
$$\frac{\partial \phi}{\partial y} = h'(y) = \frac{1+y^2}{y^3}$$
$$h(y) = \int \frac{1+y^2}{y^3} dy = -\frac{1}{2y^2} + ln|y|$$
As a result, the function is $$\phi (x,y) = \frac{x^2}{2}-\frac{1}{2y^2}+ln|y|$$

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##### Quiz-1 / TUT0201 MAT244 QUIZ 1
« on: September 27, 2019, 02:49:00 PM »
Welcome to discuss

Slove $$y'=ysinx$$
it is seperable. i.e. $$\frac{dy}{dx}=y*sinx$$
rearrange it. $$\frac{dy}{y}=sinx*dx$$
integrating both side. $$In|y|=-cosx+C_{1}$$
Finally we get: $$y = C*e^{-cosx}$$where$$C > 0$$

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