212HomeWork2sol - Physics 212A Homework#2 Name 1 a)Starting...

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Physics 212A Homework #2 Name: 1. a)Starting from the Euler relation, derive the expression cot - 1 z = 2 Log B z - z + F Euler relation is E^(I z) = Cos[z] + I Sin[z] cotz = Cos @ z D Sin @ z D TrigToExp - I ª - z + ª z M ª - z - ª z Solve @ y cotz,z D Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. :: z fi - Log B - + y - + y F> , : z fi - Log B + y - + y F>> TrigToExp @ ArcCot @ z DD 1 2 Log B 1 - z F - 1 2 Log B 1 + z F Simplify @ % D PowerExpand 1 2 H Log @ - + z D - Log @ + z DL b) Find the series expansion about z = 0 and an expansion valid for large / z / Series B - Log B + y - + y F , 8 y,0,8 <F Π 2 - y + y 3 3 - y 5 5 + y 7 7 + O @ y D 9
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s1 = Series B - Log B + y - + y F , 8 y, ¥ ,8 <F 1 y - 1 3y 3 + 1 5y 5 - 1 7y 7 + O B 1 y F 17 2 c) use this result to sum the series 1- 1 3 + 1 5 + .. and 1 - 1 3 3 + 1 5 3 3 - 1 7 3 3 3 +... . Confirm your results using Sum. Evaluating the series s1 at y -> 1 generates the sum 1- 1 3 + 1 5 + .. , s1=ArcCot[y], so the sum is ArcCot[1] ArcCot @ 1 D Π 4 The formula for the nth term is Table @H - 1 L ^ HH n LL HH 2n + 1 LL , 8 n,0,5 <D : 1, - 1 3 , 1 5 , - 1 7 , 1 9 , - 1 11 > Using Sum, we get Sum @H - 1 L ^ HH n LL HH 2n + 1 LL , 8 n,0, ¥ <D Π 4 the same as using the ArcCot series. Next, we note that the series 1 - 1 3 3 + 1 5 3 3 - 1 7 3 3 3 is 3 H s1/.y-> 3 ) , so the sum should be Sqrt @ 3 D ArcCot @ Sqrt @ 3 DD Π 2 3 Verifying with sum, the nth term is Table @H - 1 L ^ HH n LL H 3^ H n L H 2n + 1 LL , 8 n,0,6 <D : 1, - 1 9 , 1 45 , - 1 189 , 1 729 , - 1 2673 , 1 9477 > so the sum is Sum @H - 1 L ^ HH n LL H 3^ H n L H 2n + 1 LL , 8 n,0, ¥ <D 2 212HomeWork2sol.nb
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Π 2 3 2. Use Manipulate to draw the image of a unit circle centered at zc=xc+I yc under the mapping Sin[z] . xc and yc are control parameters, so you will adjust a slider to specify the location of the circle in the z plane. Use ComplexExpand and Parametric plot ( a convenient way to parametrically describe the circle is z= xc +I yc + ª Θ ) . To prevent the plot from autscaling as you adjust the parameters, it is a good idea to use PlotRange to fix the range of the plot. Manipulate @ u = ComplexExpand @ Re @ Sin @ xc + Iyc + E^ H I Θ LDDD ; v = ComplexExpand @ Im @ Sin @ xc + Iyc + E^ H I Θ LDDD ; ParametricPlot @8 u,v < , 8 Θ ,0,2 Π < ,PlotRange fi 88 - 3,3 < , 8 - 3,3 <<D , 88 xc,0 < , - 4,4 < , 88 yc,0 < , - 4,4 <D 3. Show that f[z] = Sin[z] and f[z] = ª z are differentiable, but f[z] =Conjugate[z] is not. Extract the real and imaginary parts: 8 rp,ip < = ComplexExpand @8 Re @ Sin @ x + Iy DD ,Im @ Sin @ x + Iy DD<D 8 Cosh @ y D Sin @ x D ,Cos @ x D Sinh @ y D< The Cauchy-Riemann conditions are D @ ip,y D == D @ rp,x D True - D @ rp,y D == D @ ip,x D True so the CR conditions are satisfied for S[z] The derivatives D @ rp,x D Cos @ x D Cosh @ y D D @ rp,y D Sin @ x D Sinh @ y D D @ ip,x D - Sin @ x D Sinh @ y D 212HomeWork2sol.nb 3
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D @ ip,y D Cos @ x D Cosh @ y D are also continuous everywhere, so Sin @ z D is analytic. Repeat for E^z : Extract the real and imaginary parts: 8 rp,ip < = ComplexExpand @8 Re @ E^ H x + Iy LD ,Im @ E^ H x + Iy LD<D 8 ª x Cos @ y D , ª x Sin @ y D< The Cauchy-Riemann conditions are D @ ip,y D == D @ rp,x D True - D @ rp,y D == D @ ip,x D True 8 D @ rp,x D ,D @ rp,y D ,D @ ip,x D ,D @ ip,y D< 8 ª x Cos @ y D , x Sin @ y D , ª x Sin @ y D , ª x Cos @ y D< CR is OK and derivatives are continuous, so ª z is analytic.
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