Messages

16

Term Test 1 / Re: TT1 Problem 3 (night)

« on: October 19, 2018, 10:07:07 AM »

I think $h'(x)=-2 \Rightarrow h(x)=-2x+C$, where $C$ is an arbitrary real constant.
Thus, $v(x,y)=3x^2y-y^3+3y-2x+C$.
Therefore, $f(z)=f(z)=z^3+3z-2iz+iC$

17

Term Test 1 / Re: TT1 Problem 3 (noon)

« on: October 19, 2018, 09:59:30 AM »

For part$(b)$:
CR-equation is:
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

18

Term Test 1 / Re: TT1 Problem 1 (morning)

« on: October 19, 2018, 07:50:22 AM »

$\textbf{Alternative Method}:$
$$\\$$
Since,
$$z=x+iy, \text{ where } x,y \in \mathbb{R}.$$
$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$
Then,
$$\begin{align}\cos(z)=\cos(x+iy)&=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2}\\&=
\frac{e^{-y+ix}+e^{y-ix}}{2}\\&=\frac{e^{-y}[\cos(x)+i\sin(x)]+e^{y}[\cos(-x)+i\sin(-x)]}{2}\\&=\frac{e^{-y}\cos(x)+ie^{-y}\sin(x)+e^{y}\cos(x)-e^{y}\sin(x)}{2},
\\\text{since } \cos(-x)=\cos(x) \text{ and }\sin(-x)=-\sin(x).\\&=\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2}\end{align}$$
Hence,
$$\begin{align}|\sin(z)|=|\sin(x+iy)|&=\Bigg|\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2} \Bigg|\\&=\frac{\Bigg|\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)\Bigg|}{\Big|2\Big|}\\&=\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\end{align}$$
Thus,
$$\begin{align}|\sin(z)|^2=|\sin(x+iy)|^2&=\Bigg(\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\Bigg)^2\\&=\frac{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}{4}\\&=\frac{[\cos^2(x)(e^{-2y}+e^{2y}+2)]+[\sin^2(x)(e^{-2y}+e^2y-2)]}{4}\\&=\frac{e^{-2y}(\cos^2(x)+\sin^2(x))+e^{2y}(\cos^2(x)+\sin^2(x))+2(\cos^2(x)-\sin^2(x))}{4}\\&=\frac{e^{-2y}+e^{2y}+2(2\cos^2(x)-1)}{4}\\&=\frac{e^{-2y}+e^{2y}-2+4\cos^2(x)}{4}\\&=\frac{e^{-2y}+e^{2y}-2}{4}+\frac{4\cos^2(x)}{4}\\&=\sinh^2(y)+\cos^2(x)\end{align}$$
Note:
$$\sinh(y)=\frac{e^{y}-e^{-y}}{2}, \text{ where } y\in \mathbb{R}.$$
$$\begin{align}\sinh^2(y)&=\Bigg(\frac{e^{y}-e^{-y}}{2}\Bigg)^2\\&=\frac{e^{2y}+e^{-2y}-2}{4}\end{align}$$
Therefore, we have proven $|\cos(z)|^2=\cos^2(x)+\sinh^2(y)$ for all $z=x+iy$.

19

Term Test 1 / Re: TT1 Problem 1 (morning)

« on: October 19, 2018, 07:26:54 AM »

$$\begin{align}\cos(z)=\cos(x+iy)&=\cos(x)\cos(iy)-\sin(x)\sin(iy) \\&= \cos(x)\cosh(y)-i\sin(x)\sinh(y)\end{align}$$
$$\require{cancel}\begin{align}|\cos(z)|^2=|\cos(x+iy)|^2&=\Bigg(\sqrt{\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)}\Bigg)^2 \\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y) \\&= \cos^2(x)(1+\sinh^2(x))+(1-\cos^2(x))\sinh^2(y)\\&=\cos^2(x)+\cancel{\cos^2(x)\sinh^2(x)}+\sinh^2(y)-\cancel{\cos^2(x)\sinh^2(y)}\\&=\cos^2(x)+\sinh^2(y)\end{align}$$
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20

Term Test 1 / Re: TT1 Problem 3 (night)

« on: October 19, 2018, 06:35:55 AM »

$(a).$
$$\begin{align}\frac{\partial u(x,y)}{\partial x}&=3x^2-3y^2+3\\ \frac{\partial^2u(x,y)}{\partial x^2}&=6x\ \\ \frac{\partial u(x,y)}{\partial y}&=-6xy+2\\ \frac{\partial^2u(x,y)}{\partial y^2}&=-6x\end{align}$$
Where first and second partial derivatives are continuous with respect to both $x$ and $y$.
$$\Delta u=  \frac{\partial^2u(x,y)}{\partial x^2}+ \frac{\partial^2u(x,y)}{\partial y^2}= 6x+(-6x)=0$$
Therefore, $u(x,y)=x^3-3xy^2+2y+3x$ is a harmonic function.
$\\$
$\\$
$(b).$
$\\$
Use CR-equation to find the harmonic conjugate.
$$\frac{\partial u}{\partial x }=3x^2-3y^2+3= \frac{\partial v}{\partial y }$$
$$\begin{align}\Rightarrow v(x,y)&= \int(3x^2-3y^2+3)dy \\&=3x^2y-y^3+3y+h(x) \\ \Rightarrow  \frac{\partial v}{\partial x }=6x+h'(x)\end{align}$$
Hence,$$\frac{\partial u}{\partial y } =-6xy+2=-\frac{\partial v}{\partial x }=-(6x+h'(x)) \\ \Rightarrow h'(x)=-2 \\ \Rightarrow h(x)= -2x$$
Therefore the harmonic conjugate $v(x,y)=3x^2y-y^3+3y-2x.$
$\\$
$\\$
$(c).$
$\\$
$$\begin{align}u(x,y)+iv(x,y)&=x^3-3xy^2+2y+3x+i(3x^2y-y^3+3y-2x)\\ &=x^3-3xy^2+i3x^2y-iy^3+3x+i3y+2y-i2x\end{align}$$
Consider $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
Thus $$(x+iy)^3=x^3+i3x^2y-3xy^2-iy^3=z^3$$
$$3x+i3y=3(x+iy)=3z$$
$$2y-i2x=-i2(x+iy)=-2iz$$
Therefore, $f(z)=z^3+3z-2iz$.

21

Quiz-3 / Re: Q3 TUT 0101

« on: October 12, 2018, 06:23:13 PM »

Let $$\gamma(t)=p+Re^{it}=-4+e^{it}, \text{ where } 0\leq t \leq 2\pi.$$
$$f(z)=\frac{1}{z+4}$$
Thus $$\gamma'(t)=ie^{it}$$
$$\begin{align}\int_\gamma f(z)dz&=\int_{0}^{2\pi}f(\gamma(t))\gamma'(t)dt\\&=\int_{0}^{2\pi}\frac{1}{-4+e^{it}+4}(ie^{it})dt\\&=\int_{0}^{2\pi}{e^{-it}}(ie^{it})dt\\&=\int_{0}^{2\pi}idt\\&=it\Big|_0^{2\pi}\\&=2\pi i\end{align}$$

22

Thanksgiving bonus / Re: Thanksgiving bonus 1

« on: October 06, 2018, 10:52:20 AM »

Let$$u(x,y)=\frac{x}{x^2+y^2}, v(x,y)=\frac{y}{x^2+y^2}$$
$$\frac{\partial{u}}{\partial{x}}=\frac{-(x^2-y^2)}{(x^2+y^2)^2}, \frac{\partial{u}}{\partial{y}}=\frac{2xy}{-(x^2+y^2)^2}$$
$$\frac{\partial{v}}{\partial{x}}=\frac{-2xy}{(x^2+y^2)^2}, \frac{\partial{v}}{\partial{y}}=\frac{x^2-y^2}{(x^2+y^2)^2}$$
$$\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}=0, \frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}=0, x^2+y^2>0$$
Thus,  given function represents a locally sourceless and irrotational flow.

23

Quiz-2 / Re: Q2 TUT 0301

« on: October 05, 2018, 06:39:52 PM »

By Ratio Test:
$$\begin{align}\sum_{n=1}^\infty \Biggl(\frac{1+2i}{\sqrt{6}}\Biggl)^n &= \lim_{n\to\infty}\Biggl|\frac{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n+1}}{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n}}\Biggr|\\&=\lim_{n\to\infty}\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\sqrt{\Big(\frac{1}{\sqrt{6}}\Big)^2+\Big(\frac{2}{\sqrt{6}}\Big)^2}\\&=\sqrt{\frac{5}{6}}<1\end{align}$$
Therefore,  the given infinite series converges.

24

Quiz-1 / Re: Q1: TUT 0301

« on: September 28, 2018, 05:51:56 PM »

Just notice, I posted the answer before the 18:00 you mentioned. Is this gonna be an issue? Just wondering.

25

Quiz-1 / Re: Q1: TUT 0301

« on: September 28, 2018, 04:34:22 PM »

Let $$z=x+iy,\text{where} \space x,y \in \mathbb{R}.$$Given equation becomes:
$$|x+iy-1|^2=|x+iy+1|^2+6$$
Since,$$|x+iy-1|=\sqrt{(x-1)^2+y^2} \\ |x+iy+1|=\sqrt{(x+1)^2+y^2}$$
Hence, we have$$\require{cancel}\begin{align}\bigg(\sqrt{(x-1)^2+y^2}\bigg)^2&=\bigg(\sqrt{(x+1)^2+y^2}\bigg)^2+6 \\(x-1)^2+y^2&=(x+1)^2+y^2+6 \\ \cancel{x^2}-2x+\cancel{y^2}&=\cancel{x^2}+2x+\cancel{y^2}+6\\-4x&=6\\x&=-\frac{3}{2}\end{align}$$
$\\$
$\\$
Therefore, the locus of points $z$ are the straight line(vertical line) $x=-\frac{3}{2}.$

26

MAT334--Misc / Re: What is covered in the quiz 1?

« on: September 22, 2018, 10:40:27 AM »

Since for the last two TUT, the topic of each tutorial is the same as
 the topic of the lecture in the same week instead of the previous week. I am wondering what to expect in the quiz1
Also ,for this week we covered green theorem at the end which is not mentioned in the tutorial handout at all, are we going to expect them on the quiz?

Quiz 1:
Drawn from assignments for Week 1--2; during Tutorials
http://www.math.toronto.edu/courses/mat334h1/20189/homeassignments-334.html

27

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:56:10 PM »

Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.

28

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:54:19 PM »

Small Error: $W_2(x)$ should be $2e^{4x}$.

For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

Oh, you're right. My mistake.

29

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:47:31 PM »

Small Error: $W_2(x)$ should be $2e^{4x}$.

30

Final Exam / Re: FE-P2

« on: April 11, 2018, 11:29:00 PM »

Most of the answer is correct, there is one small computational error.$\\$
For $Y_2(t)$, it should be $Y_2(t)=-4\cos(t)+2\sin(t)$.$\\$