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### Messages - Vedant Shah

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

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##### 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
##### 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:
$|z-2| = 2$

4
##### 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:
$Re[(-\frac{1}{2} + i)z]$

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##### 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.

6
##### 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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