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### Messages - hanyu Qi

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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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##### 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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##### 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.

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