### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - oighea

Pages: [1]
1
##### 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 $-l$.

2
##### 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)$are all zero. (Removable) There are no branch points on this function as it does not involve fractional powers and logarithms which are known to be multivalued. Proofs: Case 1:$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)$must have an essential nonisolated singularity at infinity. 3 ##### Term Test 1 / Re: TT1 Problem 2 (noon) « on: October 21, 2018, 02:07:22 PM » Absolute convergence: I proved$\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}$. It is also a p-series, and uses a comparison, and since the power is greater than 1, it is absolutely convergent. 4 ##### Term Test 1 / Re: TT1 Problem 2 (noon) « on: October 21, 2018, 02:52:00 AM » (a) The inequality$|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}$by De MoivrÃ©'s Theorem. Then, test absolute convergence:$\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$- the inequality isn't strict. At that value, all the terms are zero. 5 ##### Term Test 1 / Re: TT1 Problem 4 (noon) « on: October 20, 2018, 09:36:32 PM » Since the square is a piecewise simple closed curve, and that the integrand is of the form$\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$. 6 ##### Term Test 1 / Re: TT1 Problem 2 (noon) « on: October 20, 2018, 06:07:32 PM » (a) The radius is 1/3 (Ratio test).$\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. 7 ##### 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}$8 ##### Quiz-3 / Re: Q3 TUT 5301 « on: October 12, 2018, 06:52:26 PM »$\newcommand{\Arg}{\operatorname{Arg}}\newcommand{\Log}{\operatorname{Log}}$The range is "sector shaped," consisting of two ray boundaries. The first one lies at the positive Re-axis. The second ray starts at the origin and makes the angle$\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. 9 ##### 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.

Pages: [1]