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### Messages - yunhao guan

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##### MAT334--Lectures & Home Assignments / Re: 2.6 #14
« on: December 05, 2018, 10:46:47 AM »
Let f(z) = $\frac{z}{z^2+2z+5}\ ,$then $z^2+2z+5=0$. We solve the equation, and we get $z= -1 \pm 2i$, only$z= -1+ 2i$ is the upper region.
Therefore, $Res(f, -1+2i)$ = $\dfrac{\sqrt{-1+2i}}{(-1+2i)-(-1-2i)} = \frac{\sqrt{-1+2i}}{4i}\$ We need to compute $\sqrt{-1+2i}=a+ib$. We square both sides, and we get that $a = \sqrt{\frac{\sqrt{5}-1}{2}} =b^ {-1}$
Thus $I = Re(2\pi\ i$$\frac{a+ib}{4i}) = \frac{\pi\ a}{2}=\frac{\pi}{2}\sqrt{\frac{\sqrt{5}-1}{2}}$
Hope it helps.

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##### MAT334--Lectures & Home Assignments / Re: 3.2 Q6
« on: December 04, 2018, 12:22:07 AM »
Let $h = f-g$, and we know that h is analytic. Therefore $\left|e^ {f-g} \right| = e^ {Reh} = e^0 = 1$, since $Ref = Reg$
So $\left|e^ {f-g} \right| = 1$ and it implies that $e^ {f-g}$ is constant which means $e^ {f-g} = c$ so $f-g = ln(c)$