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### Topics - Xinqiao Li

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##### Chapter 3 / Confused about the notation in maximum modulus principle
« on: December 15, 2020, 10:20:52 AM »
Hello, this is the problem 5 from the sample exam 2020F, I am kind of confused by the notation here. What is meant by find instead $max_D Re(f(z))$ and $\in_D Re(f(z))$? Are we suppose to find the maximum of the real part of the function f on the bounded domain D?

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##### Quiz 2 / Quiz2 LEC0101 C
« on: October 03, 2020, 05:47:44 AM »
Questions: Find all points of continuity of the given function: $f(z)=(Imz-Rez)^{-1}$

Solutions:

Let $z=x+iy$ where x,y are real numbers.

Then $f(z)=f(x,y)=(Im(x+iy)-Re(x+iy))^{-1}=(y-x)^{-1}=\frac{1}{y-x}$

The function is not valid when the denominator equals 0, that is, $y=x$.

Therefore, the function is discontinuous only at all points of the line $y=x$.

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##### Quiz 1 / LEC0101 Quiz1
« on: September 25, 2020, 11:17:28 AM »
Problem:
Describe the locus of points z satisfying the given equation.
$|z+1|^2+2|z|^2=|z−1|^2.$

Solution:
Let $z=x+yi$ where x,y are real numbers
$|z+1|^2 = |(x+yi)+1|^2=|(x+1)+yi|^2=(x+1)^2+y^2=x^2+2x+1+y^2$
$2|z|^2=2|x+yi|^2=2(x^2+y^2)=2x^2+2y^2$
$|z-1|^2 = |(x+yi)-1|^2=|(x-1)+yi|^2=(x-1)^2+y^2=x^2-2x+1+y^2$

Hence,
$|z+1|^2+2|z|^2=|z−1|^2$
$x^2+2x+1+y^2+2x^2+2y^2=x^2-2x+1+y^2$
$2x^2+4x+2y^2=0$
$2(x^2+2x+y^2)=0$
$x^2+2x+y^2=0$
$x^2+2x+1+y^2=0$
$(x+1)^2+y^2=1$

The locus of point z is a circle centered at (-1,0) with radius 1.

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##### Quiz-4 / TUT0502 Quiz4
« on: October 18, 2019, 09:59:31 PM »
Find the general solution of the given differential equation:
$$9y'' + 6y' + y = 0$$
Solve:

The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$

Factor it we obtain $(3r + 1)(3r + 1)$

Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$

We observe repeated roots here.

Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$

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##### Quiz-3 / TUT0502 Quiz 3
« on: October 11, 2019, 02:00:35 PM »
Find the solution of the given initial problem:
$$y''+4y'+3y = 0, y(0) = 2, y'(0) = -1$$
Assume $y = e^{rt}$, then it must follow that r is the root of the characteristic polynomial
$$r^2+4r+3=0\\ (r+1)(r+3)=0$$
We have $r_1 = -1$ or $r_2 = -3$.

The general solution of the second order differential equation has the form of
$$y = c_1e^{r_1t} + c_2e^{r_2t}$$

Thus, we have
$$y = c_1e^{-t} + c_2e^{-3t}$$

The derivative of this general solution is
$$y' = -c_1e^{-t} - 3c_2e^{-3t}$$

To satisfy both initial conditions $y(0) = 2$ and $y'(0) = -1$,

We have $2 = c_1 + c_2$ and $-1 = -c_1 -3c_2$

This gives us $c_1 = \frac{5}{2}$ and $c_2 = -\frac{1}{2}$

Therefore, the solution of the initial value problem is
$$y = \frac{5}{2}e^{-t} -\frac{1}{2}e^{-3t}$$

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##### Quiz-2 / TUT0502 Quiz2
« on: October 04, 2019, 02:00:01 PM »
Determine whether the equation given below is exact. If it is exact, find the solution
$$(e^xsin(y)-2ysin(x))-(3x-e^xsin(y))y'=0\\ (e^xsin(y)-2ysin(x))dx-(3x-e^xsin(y))dy=0$$
Let $M(x,y) = e^xsin(y)-2ysin(x)$

Let $N(x,y) = -(3x-e^xsin(y)) = e^xsin(y)-3x$
$$M_y(x,y) = \frac{\partial}{\partial y} M(x,y)= \frac{\partial}{\partial y} (e^xsin(y)-2ysin(x)) = e^xcos(y)-2sin(x)\\ N_x(x,y) = \frac{\partial}{\partial x} N(x,y) = \frac{\partial}{\partial x} (e^xsin(y)-3x) = e^xsin(y)-3$$
Clearly, we see that $e^xcos(y)-2sin(x) \neq e^xsin(y)-3$

Therefore, $M_y(x,y) \neq N_x(x,y)$

By definition of exact, we can conclude the equation is not exact.

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##### Quiz-1 / TUT0502
« on: September 27, 2019, 02:08:27 PM » Pages: