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Quiz-6 / Re: Q6 TUT 0203
« on: November 18, 2018, 02:17:49 AM »
This proof does not require the direct evaluation of Laurent/Power series, but instead relies on algebraic and calculus manipulation.

Let $f(z)$ be a function analytic on the punctured disk of radius $R$ centered on $z_0$: $\{z | 0 < |z - z_0| < R\}$, and has a pole of order $l$ @ $z_0$.

Then, $\displaystyle f(z) = \frac{H(z)}{(z-z_0)^l}$, where $H$ is analytic and nonzero on all of the disk $\{z | 0 \leq |z - z_0| < R\}$.

Therefore, we express $\displaystyle H(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + ... = \sum_{k=0}a_k(z-z_0)^k$

We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(z-z_0)^l - l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = \frac{H'(z)(z-z_0)^l}{(z-z_0)^{2l}} - \frac{l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = l\frac{H(z)}{(z-z_0)^{2l - l + 1}} - \frac{H'(z)}{(z-z_0)^{2l - l}}$
We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}$

It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(z-z_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)} - l\frac{1}{(z-z_0)^1}$

From now on, we can see that the residue could be $-l$, but considering $\frac{H'(z)}{1H(z)}$, we conclude that the $H$ being analytic and nonzero on the entire disk implies $H'$ being analytic as well, and finally $\frac{H'(z)}{H(z)}$, and so that fraction has no principal part. This leaves the maximum negative degree of the principal part to be 1.

We conclude that the residue has to be $-l$.

Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P3
« on: November 04, 2018, 05:43:36 AM »
As $z^3 \tan(\pi z)\cot^2(\pi z^2)$ involves quotients of trigonometric functions, we obtain:

$\displaystyle f(z) = z^3 \frac{\sin(\pi z)}{\cos(\pi z)}\frac{\cos^2(\pi z^2)}{\sin^2(\pi z^2)}$.

Requirements for being a singular point
  • $\cos (\pi z)$ is zero, which follows $\tan (\pi z)$ is a simple pole at that point. $\cos (\theta)$ is zero where $\theta$ is a half-integer multiple of $\pi$.
  • $\sin (\pi z^2)$ is zero, which follows $\cot (\pi z^2)$ is a pole of order 2 at that point. $\sin (\theta)$ is zero where $\theta$ is an integer multiple of $\pi$.
Singular points at $\mathbb{C}$
This function is singular at all points such $\cos(\pi z)=0$ and all points such $\sin(\pi z^2)=0$.
  • Case 1: $z$ is a half-integer. Then $z = k + \frac{1}{2}, k \in \mathbb{Z}$. Then $\pi z$ will be a half-integer multiple of $\pi$. Then $\tan(\pi z)$ will have a simple pole at that point since $\sin(\pi z) \neq 0, \cos(\pi z) = 0$ (Simple pole)
  • Case 2: $z^2$ is an integer, $k$. Then $z = \sqrt{k}, k \in \mathbb{Z}$, and $z$ is either on the real or imaginary axis. Then $\pi z^2$ will be an integer multiple of $\pi$, so $\cot^2(\pi z^2)$ will have a pole up to order 2 at that point since $\sin(\pi z^2) = 0$ at that denominator.
    • Case 2a: $z^2$ is a negative integer, $-k$ where $k \in \mathbb{N}$. Then $z = i\sqrt{k}$, on the imaginary axis. Only $\sin^2(\pi z^2)$ is zero. (Double pole)
    • Case 2b: $z^2$ is a positive integer, $k$, but $z$ is irrational.  Only $\sin^2(\pi z^2)$ is zero. (Double pole)
    • Case 2c: $z$ is a positive integer as well as $z^2$. Both $\sin^2(\pi z^2)$ on denominator and $\sin(\pi z) are zero. (Simple pole)
    • Case 2d: $z$ is zero. Then $z, \sin(\pi z), \sin^2(\pi z^2)$ are all zero. (Removable)
There are no branch points on this function as it does not involve fractional powers and logarithms which are known to be multivalued.

Case 1: $z$ is a half-integer. It follows $z$ is real, and is one half more than an integer.
Then $\pi z$ is a half-integer multiple of $\pi$, so $\cos(\pi z) = 0, \sin^2(\pi z) \neq 0$.
Let $z = k + \frac{1}{2}$ such $k \in \mathbb{Z}$. Then $z^2 = (k + \frac{1}{2})^2 = [k^2 + 1] + \frac{1}{4}$, so the square of a half-integer is one quarter greater than an integer. Hence $z^2$ is neither an integer nor a half-integer multiple of $\pi$, so $\cot^2(\pi z^2) \neq 0$.

Then all the terms on the numerator are nonzero, and $\cos(\pi z) \sin(\pi z^2) = 0$ due to the zero value of the cosine. Then $\tan(\pi z)$ has a simple pole and $\csc^2 (\pi z^2)$ is nonzero.

Therefore, for all half-integer $z$, $f$ has a simple pole.

Case 2a: $z$ is an irrational real multiple of $\pi$, but $z^2$ is a positive integer.
Then $\pi z^2$ is an integer multiple of $\pi$, so $\sin^2(\pi z^2) = 0$. Also, $\pi z$ is not a rational multiple of $\pi$, so $\sin(\pi z) \neq 0, \cos(\pi z) \neq 0$. Also, note $\cos^2(\pi z^2) \neq 0$ as the argument of $\cos^2$ is an integer multiple of $\pi$, rather than a half integer multiple.

It follows $\cot^2 (\pi z^2)$ is singular, whereas $z \tan (\pi z)$ is defined. As $\cot (\pi z^2)$ has a simple pole here, it follows $\cot^2 (\pi z^2)$ has a pole of order 2.

Therefore, for all $z \notin \mathbb{Q}$ and $z^2 \in \mathbb{Z}$, $\sin^2(\pi z^2) = 0$. Then $f$ has a pole of order 2 at all points where $z^2$ is a positive integer, but the magnitude of $z$ is irrational.

Case 2b: $z$ is a real multiple of $i\pi$, and $z^2$ is a negative integer.
Then $\pi z^2$ is a negative integer multiple of $\pi$, so $\sin^2(\pi z^2) = 0$, but $\pi z$ is not a rational multiple of $\pi$, so the proof is similar to Case 2a.

Therefore, for all $z$ such $z^2$ is a negative integer, $\sin^2(\pi z^2) = 0$. Then $f$ has a pole of order 2 at all points where $z^2$ is a negative integer, but the magnitude of $z$ is irrational.

Case 2c: $z$ is a nonzero integer multiple of $\pi$.
Then $z^2$ is a positive integer multiple of $\pi$. Then $\cos(\pi z) \neq 0, \cos^2(\pi z^2) \neq 0$, but both $\sin(\pi z) = 0, \sin^2(\pi z^2) = 0$, so we have $f = z^3\frac{\cos^2(\pi z^2)}{\cos(\pi z)} \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$, where $z \frac{\cos^2(\pi z^2)}{\cos(\pi z)}$ is a defined here.

Then $g(z) = \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$ is singular where $z$ is a nonzero integer multiple of $\pi$, and that $f = z \frac{\cos^2(\pi z^2)}{\cos(\pi z)}g$.

$g(z)$ has a simple pole at nonzero integer multiples of $z$.

Therefore, $f$ has a simple pole at all nonzero integer multiples of $z$.

Case 2d: $z$ is zero.
Then $z^2$ is also an integer multiple of $\pi$, so $\cos(\pi z) \neq 0, \cos^2(\pi z^2) \neq 0$. We examine $h(z) = z \frac{\sin(\pi z)}{\sin^2(\pi z^2)}$. such $f = \frac{\cos^2(\pi z^2)}{\cos(\pi z)}h$.

We use Cauchy's Theorem to determine that $h$ and so $f$ is actually analytic on an arbitrarily small disk $D_{\delta,0} = \{z : |z| < \delta\}$.

Therefore, $f$ has a removable singularity at $z = 0$, and the limit approaches 1.

Singular points at $\infty$
Consider $\displaystyle g(z) = f(1/z) = z^{-3} \tan(\pi/z) \cot^2(\pi/z^2)$

This is an essential singularity, as $g$ approaches no limit as $z \mapsto 0$. For example, if $z$ is real and approaches positive infinity, $g$ becomes zero at points that are integer values of $\pi$ and undefined at points $z^2$ is an integer multiple of $\pi$ and nonzero elsewhere. Then it follows $f$ approaches no limit as $z \mapsto \infty$ and will contain infinitely many poles along that real axis. Hence $f(z)$ must have an essential nonisolated singularity at infinity.

MAT334--Lectures & Home Assignments / Re: 2.2 Q13
« on: October 22, 2018, 06:04:59 PM »
The textbook solution $\displaystyle z^2 \sum_{n=0}^{\infty}(-1)^n z^{2n} \left(\sum_{j=0}^n\frac{1}{(2j+1)!(2n - 2j + 1)!}\right)$ is indeed complex, even though $\displaystyle \sin^2 z = \frac{1 - \cos(2z)}{2} = \frac{1}{2} - \frac{\cos(2z)}{2}$ by using the trig ID for $\sin^2$. The author likely used the multiplication formula for $\sin^2 z$, which isn't efficient as it can be expressed as a cosine function without using the multiplcation formula. Evaluate them individually whenever you can reduce a product of functions into a more familiar form.

The power series for $\cos(z)$ is given by $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n}$, and it follows that the power series of $\cos(2z)$ is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2z)^{2n}$ Then, by scalar multiplcation by $\frac{1}{2}$, we obtain $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{2(2n)!}(2z)^{2n}$

Furthermore, express $\frac{1}{2}$ as a geometric series, $\displaystyle  \frac{1}{4}\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{4}\frac{1}{1 - 1/2} = \frac{1}{4}\frac{1}{(1/2)} = \frac{1}{4}2 = \frac{1}{2}$. Note the geometric series is $\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}, |r| < 1$.

We evaluate the power series:
$\displaystyle \sin^2 (z) = \frac{1}{2} - \frac{\cos(2z)}{2}$: The "multiplication formula" is no longer relevant. Just use scalar multiplication.
$\displaystyle = \frac{1}{2} - \sum_{n=0}^{\infty}\frac{(-1)^n}{2(2n)!}(2z)^{2n}$: Good enough, but we can do something further to deal with the constant term.
$\displaystyle = \frac{1}{2} - \frac{(-1)^0}{2(0!)}z^0 - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!} (2z)^{2n} = \frac{1}{2} - \frac{+1}{2} - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!}(2z)^{2n} = - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!} (2z)^{2n} =  + \sum_{n=1}^{\infty}\frac{(-1)(-1)^n}{2(2n)!} (2z)^{2n}$: We find out the constant term cancels out.
$\displaystyle = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2(2n)!} (2z)^{2n}$, an alternating power series with odd $n$ positive, or terms with powers divisible by 2 but not by 4 positive.

Edit: Constant term cancels out, fixed general geometric series formula.

Term Test 1 / Re: TT1 Problem 2 (noon)
« on: October 21, 2018, 02:07:22 PM »
Absolute convergence: I proved $\displaystyle= \sum_{n=0}^\infty \frac{1}{|n^2 + 1|} < \sum_{n=0}^\infty \frac{1}{|n^2|}$ is absolutely convergent for $|z| = \frac{1}{3}$. It is also a p-series, and uses a comparison, and since the power is greater than 1, it is absolutely convergent.

Term Test 1 / Re: TT1 Problem 2 (noon)
« on: October 21, 2018, 02:52:00 AM »
(a) The inequality $|z|$ being less than $\frac{1}{3}$ is not strict (Absolute Value, P-Series, Comparison)

Suppose $|z| = \frac{1}{3}$ - it is on the boundary of the radius, and $y = \mathrm{Im}\, z$.

Then we have $\displaystyle \sum_{n=0}^\infty \frac{3^nz^n}{n^2 + 1} = \sum_{n=0}^\infty \frac{3^n(|\frac{1}{3}|(\cos y + i \sin y))^n}{n^2 + 1} $

$\displaystyle= \sum_{n=0}^\infty \frac{3^n(\frac{1}{3})^n(\cos y + i \sin y)^n}{n^2 + 1}$

$\displaystyle= \sum_{n=0}^\infty \frac{1(\cos y + i \sin y)^n}{n^2 + 1} = \sum_{n=0}^\infty \frac{(\cos ny + i \sin ny)}{n^2 + 1}$ by De Moivré's Theorem.
Then, test absolute convergence:

$\displaystyle= \sum_{n=0}^\infty \frac{|(\cos ny + i \sin ny)|}{|n^2 + 1|}$

$\displaystyle= \sum_{n=0}^\infty \frac{1}{|n^2 + 1|} < \sum_{n=0}^\infty \frac{1}{|n^2|}$. Note $\displaystyle \lim_{n \to \infty}\left|\frac{1}{n^2 + 1}\right| = \lim_{n \to \infty}\left|\frac{1}{n^2}\right| = 0$

As the series is absolutely convergent when $|z| = \frac{1}{3}$, the inequality is not strict. Hence, the series converges for $|z| \leq \frac{1}{3}$ (Comparison and P-series).

(b) The series converges only when $|z| == 0$ - the inequality isn't strict. At that value, all the terms are zero.

Term Test 1 / Re: TT1 Problem 4 (noon)
« on: October 20, 2018, 09:36:32 PM »
Since the square is a piecewise simple closed curve, and that the integrand is of the form $\frac{1}{z-p}$, here $p = 0, p \in \mathrm{In}\,L$, the integral should be $i2\pi$.

We will verify by calculating the integral directly without Green's Theorem. Define $L$ as four separate smooth curves:

$\gamma_1 = (1-i)+it, \gamma'_1 = i, t \in [0,2]$
$\gamma_2 = (1+i)-t, \gamma'_2 = -1, t \in [0,2]$
$\gamma_3 = (-1+i)-it, \gamma'_3 = -i, t \in [0,2]$
$\gamma_4 = (-1-i)+t, \gamma'_4 = 1, t \in [0,2]$

The integrands will be:

$\displaystyle A_1 = \int^2_0 \frac{i}{(1-i)+it}dt = i[\frac{1}{i}i\ln(1-i+it)]^2_0 = [\ln(1-i+it)]^2_0 = \ln(1+i) - \ln(1-i) = \ln\left(\frac{1+i}{1-i}\right) = \ln\left(\frac{e^{i\pi/4}}{e^{-i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_2 = \int^2_0 \frac{-1}{(1+i)-t}dt = -[-1\ln(1+i-t)]^2_0 = [\ln(1+i-t)]^2_0 = \ln(-1+i) - \ln(1+i) = \ln\left(\frac{-1+i}{1+i}\right) = \ln\left(\frac{e^{3i\pi/4}}{e^{i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_3 = \int^2_0 \frac{-i}{(-1+i)-it}dt = -i[\frac{1}{-i}\ln(-1+i-it)]^2_0 = [\ln(-1+i-it)]^2_0 = \ln(-1-i) - \ln(-1+i) = \ln\left(\frac{-1-i}{-1+i}\right) = \ln\left(\frac{e^{-3i\pi/4}}{e^{3i\pi/4}}\right) = \ln{e^\frac{3i\pi}{2}}  = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$
$\displaystyle A_4 = \int^2_0 \frac{1}{(-1-i)+t}dt = [\ln(-1-i+t)]^2_0 = [\ln(-1-i+t)]^2_0 = \ln(1-i) = \ln(-1-i) = \ln\left(\frac{1-i}{-1-i}\right) = \ln\left(\frac{e^{-i\pi/4}}{e^{-3i\pi/4}}\right) = \ln{e^\frac{i\pi}{2}} = \frac{i\pi}{2}$

Summing the integrals of the curves $A$ together, we obtain

$A = i\pi(\frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2})=i2\pi$.

Term Test 1 / Re: TT1 Problem 2 (noon)
« on: October 20, 2018, 06:07:32 PM »
(a) The radius is 1/3 (Ratio test).

$\displaystyle \lim_{n \rightarrow \infty} \left|\frac{3^{n+1}}{(n+1)^2 + 1} z^{n+1} / \frac{3^{n}}{n^2 + 1} z^{n}\right| = \lim_{n \rightarrow \infty}\left|\frac{3^{n+1}(n^2 + 1)}{3^n((n+1)^2+1)}\frac{z^{n+1}}{z^n}\right| = \lim_{n \rightarrow \infty}\left|\frac{3(n^2 + 1)}{n^2 + 2n + 2 + 1}z\right| = 3\lim_{n \rightarrow \infty}\left|\frac{n^2 + 1}{n^2 + 2n + 3}\right||z|$.

We now have a limit of a rational function, where both sides have the same degree and leading coefficients. Hence, we have $\displaystyle 3\lim_{n \rightarrow \infty}|1z|$.

The limit now evaluates to $|3z|$. We want that to be strictly less than 1 so the series converges.

$3|z| < 1 \implies |z| < \frac{1}{3}$. Hence, the radius of convergence is $\frac{1}{3}$. Shouldn't it be $|z|$? What happens as $|z|=\frac{1}{3}$? The inequality is not strict. It converges if $|z| = \frac{1}{3}$

(b) The radius is 0. The series only converges for z == 0.

$\displaystyle \lim_{n \rightarrow \infty} \left|\frac{((n + 1)!)^3}{(2(n+1))!}z^{n+1}/\frac{(n!)^3}{(2n)!}z^n\right| = \lim_{n \rightarrow \infty} \left|\frac{((n+1)!)^3(2n)!}{(n!)^3(2(n+1))!} \frac{z^{n+1}}{z^n}\right| = \lim_{n \rightarrow \infty} \left|\frac{((n+1)!)^3(2n)!}{(n!)^3(2n + 2)!}z\right| = \lim_{n \rightarrow \infty} \left|\frac{(n!(n+1))^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}z\right| = \lim_{n \rightarrow \infty} \left|\frac{(n!)^3(n+1)^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}\right||z|$

Cancelling out factorial terms on the fraction, we obtain $\displaystyle \lim_{n \rightarrow \infty} \left|\frac{(n+1)^3}{(2n+1)(2n+2)}\right||z|$. Note that the top of the rational function in n is cubic, and the bottom is quadratic. The limit at infinity of this rational function diverges, unless z is equal to 0.

Edit: Dealt with |z| on the boundary.

Term Test 1 / Re: TT1 Problem 5 (noon)
« on: October 20, 2018, 05:27:00 PM »
The domain region is the region $\Im z \geq -\ln 2, \Re z \in [-\frac{3\pi}{4}, \frac{3\pi}{4}]$, or equivalently $y \leq \ln 2, x \leq \frac{3\pi}{4}$.

The range region is contained within the circle $|w| \leq 2$. The $\Re w \geq -|\Im w|$ can be viewed as $x \geq -|y|$. That means, it is the region to the right of line $x \geq -y$ for the top half, and $x \geq +y$ for the bottom half.

The range region must have an absolute value of Arg less or equal to $\frac{3\pi}{4}$, and a magnitude less or equal to $2$.
It can be described as $\{w | \frac{3\pi}{4} \leq \operatorname{Arg}w \leq \frac{3\pi}{4} \wedge 0 < |z| \leq 2\}$

The function $e^{iz} = e^{i(x + iy)} = e^{-y + ix}$. Hence, the x-coordinate determines the Argument, and the y-coordinate determines the magnitude (increases as y moves downwards).
Hence, the function $e^{iz} = (e^{-y})(\cos x + i \sin x)$.

Prove that this function is injective
For $\Im  z \geq -\ln 2$, it follows $-y \leq \ln 2$, and so the magnitude $e^{-y} \leq 2$.
For $-\frac{3\pi}{4} \leq \Re x \leq \frac{3\pi}{4}$, the argument of $w$ will be in $[-\frac{3\pi}{4}, \frac{3\pi}{4}]$
Two nonzero complex numbers are the same when both the magnitude and principal Argument are the same. The function $e^{iz}$ transforms the real and imaginary parts of $z$ into the argument and the reciprocal magnitude of the output. If the outputs are the same (Arg, magnitude), the inputs are the same ($x$, $-y$). Hence, this function is injective.

Prove that this function is surjective
Given any complex number with $0 < |z| \leq 2$, and $ -\frac{3\pi}{4} \leq \operatorname{Arg}w \leq \frac{3\pi}{4}$. The input $x + iy$ would have the imaginary component (encodes the reciprocal of the magnitude) to be $e^-y \leq \ln 2$, so $y \leq \ln 2$, and finally $-y \geq \ln 2$, and the real component (encodes the Argument) to be in $ -\frac{3\pi}{4} \leq x \leq \frac{3\pi}{4}$

MAT334--Lectures & Home Assignments / Re: Chapter 1.6 PG 63 ex Example 8
« on: October 16, 2018, 10:44:40 PM »
Let $a = z^2, b = 4$. By the triangle inequality, $|a + b| \leq |a| + |b|$. Therefore, by substitution, $|z^2 + 4| < |z^2| + 4 < |z^2| - 4$, and note $|z^2| == |z|^2$.

To visualize the triangle inequality in the complex plane, the lengths of the two sides of the origin are given as $|a|, |b|$, and the third side is $|a + b|$. As expected, the third side is shorter than the sum of the two sides.

The complex triangle inequality can be proved by squaring.

There is no typo. Also, we can view $u$ as $\mathfrak{Re}f$ $v$ as $\mathfrak{Im}f$. The equation in 2.1 is the total derivative, whereas the equation in 1.6 are the two partial derivatives, one in respect to the real values $\frac{\delta u}{\delta x} + i\frac{\delta v}{\delta x}$ and the other in respect to the imaginary values $\frac{\delta u}{i\delta y} + \frac{+\delta v}{\delta y}$ (note that in the limit definition, you have to watch out for the $i$ on the denominator $k$).

The total derivative can be expressed as $\frac{df}{dz} = \frac{\delta u}{\delta x}\frac{dx}{dz} + i\frac{\delta v}{\delta x}\frac{dx}{dz} - i\frac{\delta u}{\delta y}\frac{dy}{dz} + \frac{\delta v}{\delta y}\frac{dy}{dz}$

MAT334--Lectures & Home Assignments / Re: Section 2.1 Question 1
« on: October 15, 2018, 07:00:48 PM »
If the question asks you to use the limit definition, use the limit definition.

The sin function is actually derived from the exponential function. We know the natural exponential function is differentiable, and is equal to its own derivative.

$\sin z = \frac{e^{iz} - e^{-iz}}{2i}$
$(\sin z)' = \frac{ie^{iz} + ie^{-iz}}{2i} = i \frac{e^{iz} + e^{-iz}}{2i} = \frac{e^{iz} + e^{-iz}}{2} = \cos z$

Use the limit definition of $e^z$ and the property of limits when regarding addition, multiplication, and composition to obtain the derivative of $\sin$

Quiz-3 / Re: Q3 TUT 5301
« on: October 12, 2018, 06:52:26 PM »
The range is "sector shaped," consisting of two ray boundaries. The first one lies at the positive Re-axis. The second ray starts at the origin and makes the angle $\alpha\pi$ from the first, counterclockwise. Fill in the region bounded by these two rays starting from the positive Re-axis and continue counterclockwise, and dot the border.

The range will never "overlap" as $0 < \alpha < 2$, so the possible principal arguments of $z^\alpha$ is always $0 < \Arg (z) < 2\pi$.

The appropriate choice for $\log z$ is $\Log  z$.

Proving $f(z) = z^\alpha$ maps one-to-one to the sector range: $f(z_1) = f(z_2) \Rightarrow z_1 = z_2$.
Let $z = re^{i\theta}$, such that $r$ is the magnitude, and $\theta$ is the principal argument. Note $0 < \theta < \pi$, the inequalities are strict.
Then $z^\alpha = r^\alpha e^{i\alpha\theta}$. Note $0 < \Arg (z^\alpha) < \alpha\pi$
Since $f(z_1) = f(z_2) = r^\alpha e^{i\alpha\theta}$, it follows that $r$ and $\theta$ for both $z_1, z_2$ are the same.
We conclude $z_1 = z_2 = re^{i\theta}$, so $f(z)$ is injective.

Proving $f(z) = z^\alpha$ maps onto the range: For all $f(z)$ on the "range" domain ${w: 0 < \Arg  w < \alpha\pi}$.
Since $f(z) = r^\alpha e^{i\alpha\theta}$, it follows $z = re^{i\theta}$.
We note that $0 < \Arg f(z) < \alpha\pi$, and it follows that $0 < \Arg z < \pi$.
We conclude that $f(z)$ maps the upper half-plane onto all of the range domain.

Proving that $f$ also carries the boundary to the boundary:
The boundary of the domain domain consists of all the real numbers.
The boundary of the range domain consists of 0, the positive Re-axis, and the ray $\alpha\pi$ from that axis counterclockwise, which is $\{w: \Arg  w = \alpha\pi\}$.
If $z$ is zero, $z^\alpha$ is also zero.
If $z$ is positive real, $z^\alpha$ is also positive real, with magnitude raised to the power of $\alpha$. $\Arg  z = \Arg  z^\alpha = 0$.
If $z$ is negative real, $\Arg  z = \pi$, and it follows $\Arg  z^\alpha = \alpha\pi$.The domain domain is the open upper-half plane, the set of all the all the complex imaginary numbers with positive imaginary part. Fill the upper half of the plane and dot the border.

Quiz-3 / Re: Q3 TUT 0203
« on: October 12, 2018, 06:23:43 PM »
Please note that the curve is a circle of radius 2 centered on z=0. The parametric equation of the curve is $\gamma(t) = 2e^{it}$, and the derivative is $\gamma'(t) = 2ie^{it}$. Otherwise, it is mostly correct as it is a line integral of a simple-closed curve around a holomorphic function $f(z) = z^2 + 3z + 4$. Otherwise, your answer and steps are mostly correct.

Also note that the integrand $\int_a^bf(\gamma(t))\gamma'(t)\,dt$ is of the form when you differentiate a function composition. Hence, as a shorthand, you can calculate or verify your answer as $F(\gamma(b)) - F(\gamma(a))$

To verify, $F(z) = \frac{1}{3}z^3 + \frac{3}{2}z^2 + 4z$.
Then $F(\gamma(z)) = \frac{1}{3}(2e^{it})^3 + \frac{3}{2}(2e^{it})^2 + 4(2e^{it})$
$F(\gamma(z)) = \frac{8}{3}e^{3it} + \frac{12}{2}e^{2it} + 8e^{it}$, and $f(\gamma(z))\gamma'(z) = (2^2e^{2it} + 3\cdot2e^{it} + 4)(2ie^{it}) = 8ie^{3it} + 12ie^{2it} + 8ie^{it}$
$F(\gamma(z)) = \frac{8}{3}e^{3it} + 6e^{2it} + 8e^{it}$.
The integral is then expressed as $F(2\pi) - F(0)$.
We found out that $e^{2it} = e^{4it} = e^{6it} = e^{kit} = e^0 = 1$ for all $k \in \mathbb{Z}$.
Hence \frac{8}{3}e^{3it} + 6e^{2it} + 8e^{it} = \frac{8}{3} + 6 +8$ when $t = 0$ and $t = 2\pi$.
As $F(2\pi) = $F(0)$, it follows the integral is zero.

MAT334--Lectures & Home Assignments / Re: Section 1.6 Question 3
« on: October 09, 2018, 06:28:56 PM »
You do not need to explicitly declare the components $x, y$ nor use $z\bar{z}$, just express them in terms of $t$ and solve using antiderivatives. Instead, convert $z$ into the parametric curve function $\gamma(t)$ so you have an integral in respect to $t$ quite familiar in calculus. To deal with the magnitude term, identify the real and imaginary components inside the absolute value bars.

Here, parameterize the curve $\gamma(t) = [2+t] + [it]$ where $t \in [0, 1]$. Note that the the number begins at $2 + 0i$ and both components increase by 1.

Hence, $\gamma(t) = t + 2 + it$, $f(\gamma) = |\gamma|^2$, and $\gamma'(t) = 1 + i$.

Then, we therefore have the integral $\int_0^1 f(\gamma(t))\gamma'(t) \:dt = \int_0^1|t + 2 + it|^2(1 + i) \:dt$, which will be quite familiar in earlier calculus courses. You can proceed easily from here after expressing the magnitude in terms of $t$.

Furthermore, note that this line integral is expressible as $[F(\gamma(t))]^1_0$ where $F$ is the antiderivative of $f$, since the integrand is of the form of the derivative of a function composition (that will be useful for similar line integral problems).

$\int_0^1 |[t + 2] + [it]|^2(1 + i) \:dt$
$= (1 + i)\int_0^1 |[t + 2] + [it]|^2 \:dt$, {$x=[t+2], y=[t]$}
$= (1 + i)\int_0^1 (t+2)^2 + (t)^2 \:dt$, {definition of magnitude}
$= (1 + i)\int_0^1 t^2 + 4t + 4 + t^2 \:dt$
$= (1 + i)\int_0^1 2t^2 + 4t + 4 \:dt$
$= (1 + i)[\frac{2}{3}t^3 + 2t^2 + 4t]^1_0$
$= (1 + i)[\frac{2}{3}(1) + 2(1) + 4(1)] - 0$
$= (1 + i)[\frac{2}{3} + 6]$
$= (1 + i)[\frac{20}{3}]$

MAT334--Lectures & Home Assignments / Re: Section 1.4 Question 12
« on: October 04, 2018, 01:47:33 AM »
Here is a major hint: log in some contexts may actually mean ln when the variables is real. Also, note that the absolute value |.| maps from $\mathbb{C}\mapsto \mathbb{R}^+$. Hence, log|z-2| is just a real logarithm, and is in fact defined everywhere on the complex plane except for z=2. This can be done without using the delta-epsilon definition of limits (which is unfortunately hard to verify).

We have the expression $\lim\limits_{z \to 2} (z-2) \ln|z-2|$. The lim is of the form $0\cdot-\infty$, meaning L'Hôpital's Rule applies.

Note that for L'Hôpital's Rule, $0 = \frac{1}{\infty}$. Hence, $(z-2) = \frac{1}{(z-2)^{-1}}$, and the lim of both when they approach 2 are zero.

$\lim\limits_{z \to 2} (z-2) \ln|z-2| = \lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}}$, which is of the $\frac{\infty}{\infty}$ form, the desired indeterminate form.

We apply L'Hôpital's Rule: $\lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}} \Rightarrow \lim\limits_{z \to 2}\frac{(z-2)^{-1}}{-1(z-2)^{-2}}=\lim\limits_{z \to 2}-\frac{(z-2)^{-1}}{(z-2)^{-2}}$. That looks very good, except the exponents are both negative, meaning that they switch places on the fraction bar.

We now arrive at $\lim\limits_{z \to 2}-\frac{(z-2)^{2}}{(z-2)^{1}}$, and finally $\lim\limits_{z \to 2}(z-2)$. We evaluate the limit directly: $(2-2) = 0$

∴ the lim of $(z-2) \ln|z-2|$ is zero.

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