Q4 TUT 0203

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}$
$\renewcommand{\Im}{\operatorname{Im}}$
Evaluate the given integral using the technique of Example 10 of Section 2.3 of the Textbook;
indicate which theorem or result you used to obtain your answer.
$$
\int_\gamma \Bigl(z+\frac{1}{z}\Bigr)\,dz,
$$
where $\gamma$ is any curve in $\{z\colon \Im z>0\}$ joining $(-4+i)$ to $(6+2i)$.

[Jeffery Mcbride]

\begin{equation*}
f( z) \ =\ z\ +\ \frac{1}{z} \ is\ the\ derivative\ of\ F( z) \ =\ \frac{z^{2}}{2} \ +\ Log( z)\\
\\
This\ is\ valid\ only\ where\ the\ function\ Log( z) \ is\ analytic.\ This\ domain\ in\ cludes\ the\ domain\ \\
Im\ z\  >\ 0.\ Hence,\\
\\
\int _{\gamma }\left( z\ +\ \frac{1}{z} \ \right) dz\ =\ \int _{\gamma } f( z) dz\ =\ \ \int _{\gamma } F'( z) dz\\
\\
=\ F( endpoint) \ -\ F( initial\ point)\\
\\
=F( 6+2i) \ -\ F( -4\ +\ i)\\
\\
=\left(\frac{( 6+2i)^{2}}{2} \ +\ Log( 6\ +\ 2i)\right) \ -\ \left(\frac{( -4+i)^{2}}{2} \ +\ Log( -4\ +\ i)\right)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ Log( 6\ +\ 2i) \ -\ Log( -4\ +\ i)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ \left( log\left(\sqrt{40}\right) \ +\ iArg( 6\ +\ 2i)\right) \ -\ \left( log\left(\sqrt{17}\right) \ +\ iArg( -4\ +\ i)\right)
\end{equation*}

[Victor Ivrii]

Could you write the Arg-part in terms of inverse trigonometric functions?

[Jeffery Mcbride]

Like so?


\begin{equation*}
f( z) \ =\ z\ +\ \frac{1}{z} \ is\ the\ derivative\ of\ F( z) \ =\ \frac{z^{2}}{2} \ +\ Log( z)\\
\\
This\ is\ valid\ only\ where\ the\ function\ Log( z) \ is\ analytic.\ This\ domain\ includes\ the\ domain\ \\
Im\ z\  >\ 0.\ Hence,\\
\\
\int _{\gamma }\left( z\ +\ \frac{1}{z} \ \right) dz\ =\ \int _{\gamma } f( z) dz\ =\ \ \int _{\gamma } F'( z) dz\\
\\
=\ F( endpoint) \ -\ F( initial\ point)\\
\\
=F( 6+2i) \ -\ F( -4\ +\ i)\\
\\
=\left(\frac{( 6+2i)^{2}}{2} \ +\ Log( 6\ +\ 2i)\right) \ -\ \left(\frac{( -4+i)^{2}}{2} \ +\ Log( -4\ +\ i)\right)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ Log( 6\ +\ 2i) \ -\ Log( -4\ +\ i)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ \left( log\left(\sqrt{40}\right) \ +\ iArg( 6\ +\ 2i)\right) \ -\ \left( log\left(\sqrt{17}\right) \ +\ iArg( -4\ +\ i)\right)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ log\left(\sqrt{40}\right) \ \ -\ log\left(\sqrt{17}\right) \ +\ i\ tan^{-1}\left(\frac{1}{3}\right) \ -\ i\ \ \left( \pi \ -\ tan^{-1}\frac{1}{4}\right)
\end{equation*}