Messages

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MAT334--Lectures & Home Assignments / FE Sample Question 4 (a)

« on: December 07, 2018, 05:28:43 PM »

Hello, I am wondering whether we can have different function f(z) for this question.

In the posted solution, Yilin set $ f(z) = \lambda \frac{z-a}{1-\bar{a}z}$ $| \lambda | = 1 $ and get final result as $ f(z) = \frac{5+z}{1+5z}$.

Based on a hint in Textbook 3.3 Example 1, I try to set $ f(z) = \lambda \frac{a-z}{1-\bar{a}z}$ and $| \lambda |=1 $then I did the following computation.

let $\lambda = e^{it}$ , $ a = re^{i \theta}$

$f(0) = 5$ -> $\lambda a = 5$ and $ |\lambda a| = |a| = 5 $ so $ a = 5e^{i\theta}$

$\lambda a = 5e^{it} e^{i\theta} =5$ so $e^{-it} = e^{i\theta}$

$ f(-1) = \lambda \frac{a+1}{1+\bar{a}} = -1$ so $e^{-i\theta} = -1$ ---> $e^{it} =1$ and $ \theta = \pi $

so $\lambda = 1$ and $a = -5$

$f(z) = \frac{-5-z}{1-5z}$

I am not sure if there is a computation mistake for my solution or we could set f(z) in many forms.

Thank you.

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MAT334--Lectures & Home Assignments / Re: TT2 Q4 Question for step4

« on: December 02, 2018, 01:41:20 PM »

Never mind. I think I know why.

In the answer, f(z) is integral over $-\gamma_{\epsilon}$. I guess this is why its range is $[0,\pi]$

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MAT334--Lectures & Home Assignments / TT2 Q4 Question for step4

« on: December 02, 2018, 01:36:30 PM »

Hello everyone, I am wondering why the range of $\theta$ is $[0,\pi]$ instead of $[\pi,0]$.

Then the integral estimation would be $ |\int_{\gamma_{\epsilon}} f(z) \text{d}z| \leq \int_{\pi}^{0} |f(z)| \text{d}z = \frac{-\pi \epsilon}{\sqrt{\epsilon} (1-{\epsilon}^2)}$ goes to 0 as $\epsilon$ close to 0+.

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End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 2

« on: December 01, 2018, 09:33:23 PM »

(b)

when $ x = p$

$ u(b,y) = \cos b \cosh y = A \cosh y $

$ v(b,y) = -\sin b \sinh y = B \sinh y $

$ \frac{u}{A} = \cosh y $

$ \frac{v}{B} = \sinh y $

$ \frac{u^2}{A^2} - \frac{v^2}{B^2} = (\cosh y)^2 - (\sinh y)^2  = 1 $ with $ A^2 + B^2 = (\cos b)^2 + (\sin b)^2 = 1 $

$A = \cos b$

$B = -\sin b $

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End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 2

« on: December 01, 2018, 09:17:43 PM »

(a)

$ \cos z = (\cos x)(\cosh y) - i(\sin x)(\sinh y) $

$ u(x,y) = (\cos x)(\cosh y)     v(x,y) = -(\sin x)(\sinh y) $

$ when y =q $

$ u(x,q) = (\cos x)(\cosh q) = a(\cos x) $

$ v(x,q) = -(\sin x)(\sinh q) = b(\sin x) $

$ \frac{u^2}{a^2} = (cosx)^2 $         $ \frac{v^2}{b^2} = (sinx)^2 $

$ \frac{u^2}{a^2} + \frac{v^2}{b^2} =  (cosx)^2 + (sinx)^2 = 1 $

So, lines {z: Imz = q} are mapped onto confocal ellipses {w=u+iv: $ \frac{u^2}{a^2} + \frac{v^2}{b^2} = 1 $} with $ a^2 - b^2 = 1 $ since $ (coshq)^2 - (-sinhq)^2 = 1 $

$ a = \cosh q $
$ b = -\sinh q $

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MAT334--Lectures & Home Assignments / Re: 2.6 Q14

« on: December 01, 2018, 06:53:08 PM »

Consider $ f(z) = \frac{\sqrt z}{z^2 + 2z +5}$

$z^2 + 2z +5 \implies $z= -1+2i or -1-i2.

Only z=-1+2i is up.

Res(f,-1+2i) = $ \frac{\sqrt {-1+2i}}{(-1+2i)-(-1-2i)} $= $\frac{\sqrt {-1+2i}}{4i}$

We compute $ \sqrt{-1+2i}$ = a+ib. There are two solutions, but we must choose only the one whose argument is hafl of the grgument of -1+2i.

$a^2$ - $b^2$ = -1  ab = 1

a = $ \sqrt{\frac{\sqrt5 -1}{2}} $ = $ \frac{1}{b} $

I = re( $2 \pi i \frac{a+ib}{4i})$ = $ \frac{\pi}{2}$ $ \sqrt{\frac{\sqrt5 -1}{2}}$

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Term Test 2 / Re: TT2B Problem 4

« on: December 01, 2018, 05:10:08 PM »

(a)

By residue thrm, $$ \int_{\gamma_{R,\epsilon}} f(z) \text{d}z = 2\pi i Res(f(z),i) = \lim_{z \rightarrow i} (z-i) f(z) = \frac{\ln i \sqrt i}{2i} = \frac{\pi \sqrt i }{4} = \frac{ \sqrt 2 \pi}{8} + i \frac{\sqrt 2 \pi}{8} $$

Since $$ \sqrt i = \frac{\sqrt 2 }{2} + \frac{\sqrt 2 i}{2} $$

$$ \ln i = \ln |i| + i arg(i) = i \frac{ \pi }{2} $$

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Term Test 2 / Re: TT2B Problem 1

« on: November 24, 2018, 05:37:34 PM »

just want to add a little detail on (c), if both points are inside we can use residue theorem so we can get 2πI ( res(f,3+4i)+res(f,3-4i)).

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Term Test 2 / Re: TT2 Problem 2

« on: November 24, 2018, 05:33:36 PM »

Can we use geometric series on f(z) and assume |z|<1, then we can write 1/√(1-z) directly into Laurent series.

How?

 However the series decomposition may be different but the radius of convergence is the same.

We are not looking for just radius. Basically this post was a flood. V.I.

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MAT334--Lectures & Home Assignments / Section 2.1 Question 1

« on: October 15, 2018, 04:50:02 PM »

Hello everyone,

I am wondering whether I need to show a function is differentible first then calculate the derivative even if the question only ask me to establish the derivative.

E.X.  (sinZ)’ = cosZ

The answer uses the limit definition to show but can I use theorem 3 (C-R equations) to show it as well?

Thank you!

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Quiz-2 / Re: Q2 TUT 5201

« on: October 08, 2018, 09:26:37 PM »

In attachment.

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Quiz-2 / Re: Q2 TUT 5301

« on: October 08, 2018, 09:10:52 PM »

In attachment.