Author Topic: Q4 TUT 5201  (Read 515 times)

Victor Ivrii

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Q4 TUT 5201
« on: October 26, 2018, 05:55:36 PM »
\begin{problem}
Evaluate the given integral using Cauchy’s Formula or Theorem. Orientation counter-clockwise:
$$
\int_{|z|=2} \frac{e^z\,dz} {z(z-3)}.
$$
« Last Edit: October 27, 2018, 01:30:59 PM by Victor Ivrii »

ZhenDi Pan

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Re: Q4 TUT 5102
« Reply #1 on: October 26, 2018, 05:59:31 PM »
First we have
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz
\end{equation}
The point $3$ is outside of the circle $\mid z \mid = 2$ and the point 0 is inside of the circle $\mid z \mid = 2$. Hence
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz  = \int\limits_{\mid z \mid = 2}\frac{\frac{e^z}{z-3}}{z}\,dz
\end{equation}
This gives us function $f(z)$
\begin{equation}
f(z) = \frac{e^z}{z-3} \Rightarrow f(0) = -\frac{1}{3}
\end{equation}
So Cauchy's Formula gives us
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz =2\pi i\frac{e^0}{0-3}= -\frac{2\pi i}{3}
\end{equation}

See the attachment for the circle in this question.
« Last Edit: October 27, 2018, 03:47:34 AM by ZhenDi Pan »

Jeffery Mcbride

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Re: Q4 TUT 5102
« Reply #2 on: October 26, 2018, 06:00:06 PM »

\begin{equation*}
\int _{|z|\ =\ 2} \ \frac{e^{z}}{ \begin{array}{l}
z( z-3)\\
\end{array}}\\
\\
=\ \int _{|z|\ =\ 2} \ \frac{e^{z}}{ \begin{array}{l}
( z-0)( z-3)\\
\end{array}}\\
\\
=\ \int _{|z|\ =\ 2} \ \frac{\zeta ( z)}{ \begin{array}{l}
( z-0)\\
\end{array}} \ ,\ \zeta ( z) \ =\ \frac{e^{z}}{( z-3)}\\
\\
=( 2\pi i) \zeta ( 0) \ =\ 2\pi i\ \frac{e^{0}}{( 0\ -\ 3)} \ =\ \frac{-2\pi i}{3}
\end{equation*}