Messages

1

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

« on: October 19, 2018, 09:23:34 AM »

(a)
$
U_{xx} = \frac{\partial}{\partial x} \frac{\partial}{\partial x} U \\
= \frac{\partial}{\partial x} 8y^3 -24x^2 y +5 \\
= -48xy \\
\\
U_{yy} = \frac{\partial}{\partial y} \frac{\partial}{\partial y} U \\
= \frac{\partial}{\partial y} 24x y^2 - 8x^3 \\
= 48xy \\
U_{xx} + U_{yy} = -48xy + 48xy = 0
$
Thus, U is harmonic.

(b)
By Cauchy Reimann:
$
V_y = U_x = 8y^3 - 24x^2 y + 5
\Rightarrow V = 2y^4 - 12 x^2 y^2 +5y +h(x)\\
V_x = -U_y = -24x y^2 + 8x^3
\Rightarrow V = -12 x^2 y^2 + 2x^4 + g(y) \\
\Rightarrow V(x,y) = 2x^4 - 12 x^2 y^2 + 2y^4 + 5y
$

(c)
$$
f(x,y) = U(x,y) + iV(x,y) = 8xy^3 - 8x^3y+5x + i(2x^4 - 12 x^2 y^2 + 2y^4 + 5y) \\
f(x,y) = 2i(x^4 + 4ix^3y - 6x^2y^2 -4ixy^3 +y^4) + 5(x+iy) \\
f(x,y) = 2i(x+iy)^4 + 5(x+iy) \\
f(z) = 2i ({z}) ^4 + 5z\color{red}{+Ci}.
$$

2

Term Test 1 / Re: TT1 Problem 2 (morning)

« on: October 19, 2018, 08:59:45 AM »

(a)
By Ratio Test:
$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{2^{n+1} z^{n+1} {n+1}}{3^{n+1}} \frac{3^{n}}{2^{n} z^{n} n}| \\
=\lim_{n \to \infty}  |\frac{n+1}{n} \frac{2}{3} z| \\
= |\frac{2}{3} z|
$

We want $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1$:
$
|\frac{2}{3} z| < 1 \\
|z| < \frac{3}{2} \\
R = \frac{3}{2}
$

Now testing $|z| = \frac{3}{2}$:
$
\sum\limits_{n=1}^{\infty} |\frac{2^{n} z^{n} {n}}{3^{n}}| \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |z|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |\frac{3}{2}|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} |n|
$

The series diverges at the boundary, thus we have:
$|z| < \frac{3}{2}$

(b)
By Ratio Test:
$
\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{ {(n+1)}! z^{n+1} }{{(2n + 2)}!} \frac{{(2n)}!}{{n}! z^{n}}| \\
=\lim_{n \to \infty}  |\frac{n+1}{(2n+1)(2n+2)}z| \\
= 0
$

Thus, $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1 \forall z$
Therefore, we have the series converges for:
$
R = \infty \\
|z| < \infty
$

3

MAT334--Misc / Class Participation

« on: October 08, 2018, 04:30:27 PM »

How is the Class and Tutorial participation calculated for the bonus component of the final grade?

4

Quiz-1 / Re: Q1: TUT 0203

« on: September 28, 2018, 06:03:17 PM »

This is the circle centered at $z_0=2$ with radius 2.
In other words, it is the set of points 2 units away from $z_0 = 2$. The distance of a given point, $z$, from $z_0$ is:
$d=|z-z_0|$
Thus the equation of this circle is:
$|z-2| = 2$

5

Quiz-1 / Re: Q1: TUT 0201 and TU 0202

« on: September 28, 2018, 06:01:36 PM »

Let $p = -1+2i$ and $q=1-2i$
The perpendicular bisector is the set of points equidistant to points $p$ and $q$. The distance between some point, $z$ and $p$ is $|z-p|$. Similarly, The distance between some point, $z$ and $q$ is $|z-q|$.
Thus the set of points equidistant to both $p$ and $q$ is given by the equation:
$|z-p| = |z-q|$

Another way to do this is solving for the equation of the line in x-y coordinates first. The points are $(-1, 2) and (1, -2)$, the midpoint is $(0, 0)$ and the slope is $m_1=-2$. Thus the perpendicular bisector has the slope $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$. Therefore, the perpendicular bisector has the equation:
$Re[(-\frac{1}{2} + i)z]$

6

Quiz-1 / Re: Q1: TUT 0101 and TUT 0102

« on: September 28, 2018, 06:01:05 PM »

Let $z=x+iy$
Then $Re(z) = x$ and $|z-i| = |x+i(y-1)|$
Thus:
$Re(z) = |z-i|$
$x = |x+i(y-1)|$
$x = \sqrt{x^2 + (y-1)^2}$
$x^2 = x^2 +(y-1)^2, x \ge 0$
$(y-1)^2 = 0, x \ge 0$
$y=1, x \ge 0$

In complex terms:
$ y = 1 \iff Im(z) = 1$ and $ x \ge 0 \iff Re(z) \ge 0$

Thus the equation of the line in complex terms is:
$Im(z) = 1, Re(z) \ge 0$

This is the horizontal half line extending from $z=i$ rightward.

7

Quiz-1 / Re: Q1: TUT 5301

« on: September 28, 2018, 06:00:36 PM »

Let the lines $Re(a+ib)=0$ and $Re(c+id)=0$ be perpendicular.
From section 1.2: Let $a = A+iB$ and $c= C+iD$. Then the lines are $Ax-By+Re(b)=0$ and $Cx-Dy+Re(d)=0$
Setting the slope of the first equal to the negative reciprocal of the other we get: $\frac{A}{B} = - \frac{D}{C} \iff AC=-BD$
Finally, $Re(a \bar{c}) = Re[(A+iB)(C-iD)]=AC+BD=-BD+BD=0$

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MAT334--Lectures & Home Assignments / Re: To show z1z2 = 0 implies z1 = 0 or z2 = 0

« on: September 24, 2018, 07:10:07 PM »

Given:
$ z_1 z_2 = 0$
$\Rightarrow |z_1 z_2| = |0|$
$\Rightarrow|z_1||z_2| = 0$
$\Rightarrow|z_1| = 0 \vee |z_2| = 0$
We know (Modulus is non-negative; Positive for non-zero numbers):
$|z| = 0 \iff z=0$
Thus:
$ z_1 z_2 = 0 \iff |z_1| = 0 \vee |z_2| = 0 \iff z_1 = 0 \vee z_2 = 0$

Another method would be to use polar coordinates:
$z_1 z_2 = r_1 e^{i \theta _1} r_2 e^{i \theta _2}$
$ 0 =r_1 r_2 e^{i (\theta _1 + \theta _2)}$
Since $e^{z} \gt 0$ $\forall z$:
$ 0 = r_1 r_2 \iff r_1 = 0 \vee r_2 = 0 \iff z_1 = 0 \vee z_2 = 0$

I am sure using rectangular would also work, but it if probably more work and algebra than necessary.

9

MAT334--Lectures & Home Assignments / Re: Section 1.4 Example 8

« on: September 24, 2018, 11:22:43 AM »

Start with:
$-1 \le \sin(\frac{n \pi}{4}) \le 1$
$\frac{-1}{n} \le \frac{\sin(\frac{n \pi}{4})}{n} \le \frac{1}{n} $
$\lim_{n \to \infty} \frac{-1}{n} \le \lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le \lim_{n \to \infty} \frac{1}{n}$
$ 0 \le lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} \le 0$

Thus by squeeze theorem,
$\lim_{n \to \infty} \frac{\sin(\frac{n \pi}{4})}{n} = 0$

After a similar argument, we get that:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0$

We also have:
$lim_{n \to \infty} i = i$

Since all three limits exist, we have:
$lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + i \frac{sin(\frac{n \pi}{4})}{n} = lim_{n \to \infty} \frac{cos(\frac{n \pi}{4})}{n} + lim_{n \to \infty} i *lim_{n \to \infty} \frac{sin(\frac{n \pi}{4})}{n} = 0+i0 = 0$

10

MAT334--Lectures & Home Assignments / Re: Section 1.4 Example 8

« on: September 23, 2018, 05:25:09 PM »

Can we use squeeze theorem on $\cos{\frac{n \pi}{4}}$ and $ i \sin{\frac{n \pi}{4}}$?
Then the sum of the limits is the limit of the sums (provided they both exist)

11

MAT334--Lectures & Home Assignments / Section 1.2 Question 18

« on: September 23, 2018, 05:19:33 PM »

I'm struggling with this question, and I was hoping someone could help me out: $\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}$


Show that two lines $\Re(a+ib)=0$ and $\Re(c+id)=0$ are perpendicular  $ \iff \Re(a \bar{c}) = 0$
From section 1.2: Let $a = A+iB$ and $c= C+iD$. Then the lines are $Ax-By+\Re(b)=0$ and $Cx-Dy+\Re(d)=0$
Setting the slope of the first equal to the negative reciprocal of the other I get: $\frac{A}{B} = - \frac{D}{C} \iff AC=-BD$
Finally, $\Re(a \bar{c}) = AC-BD= 2AC$

How do I proceed?

Thanks!

12

MAT334--Lectures & Home Assignments / Re: Section 1.2 question 4

« on: September 20, 2018, 11:55:52 AM »

$\{x+iy\colon x=c\}$ is not a point the plane, it is a set of points (a vertical line).

13

MAT334--Misc / Re: Difference Between Lecture Sections

« on: September 20, 2018, 11:52:11 AM »

Will we be writing the same exams and midterms?