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MAT334--Lectures & Home Assignments / Re: 2.6 #4

« on: December 07, 2018, 12:13:10 AM »

We consider the function $f(z)=\frac{e^{iaz}}{(z^2+1)(z^2+4)}$.
$(z^2+1)(z^2+4)=0\Rightarrow$$z=\pm i,$$\pm 2i.$ Only i, 2i are in the upper-half plane.
$Res(f,i)=\frac{e^{iai}}{2i(-1+4)}=\frac{e^{-a}}{6i}$    $Res(f,2i)=\frac{e^{ia2i}}{(-4+1)4i}=\frac{e^{-2a}}{-12i}$. $I=2\pi i(\frac{e^{-a}}{6i}-\frac{e^{-2a}}{12i})=\pi(\frac{e^{-a}}{3}-\frac{e^{-2a}}{6})$

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MAT334--Lectures & Home Assignments / Re: 3.2 Q7

« on: December 04, 2018, 11:26:11 PM »

The function $h(z)=e^{F(z)}$ is analytic on the disc $\{\mid$$z$-$z_0\mid$$\leq$$r\}$, it never equals zero, and $\mid$$h(z_0)\mid=1$.
Hence the maximum and the minimum are attained on the boundary circle $\{\mid$$z$-$z_0\mid$$=r\}$.
Let $z_{max}$, $z_{min}$ be the corresponding points, so $1<\mid$$h(z_{max})\mid$=$e^{Re(z_{max})}$, $1>\mid$$h(z_{min})\mid$=$e^{Re(z_{min})}$.
We deduce that Re($f(z_{max})$)$>0>Re(f(z_{min}))$

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MAT334--Lectures & Home Assignments / 3.2 Q6

« on: December 03, 2018, 11:52:10 PM »

Can someone help me solve this problem?