Toronto Math Forum

Never got any feedback on this answer, is it correct?

You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \  >\ 0$ 

When $\displaystyle z\ \rightarrow \ \pm \infty $, Re(f(z)) $\displaystyle \rightarrow \ \infty $ and Im(f(z)) $\displaystyle \rightarrow \ \infty \ $and the endpoints of $\displaystyle f([ -R,\ R]) \ $lie in the first quadrant for R really big. So the change in argument is going to be 0.

This can be coupled with what I said about the curve.

$\displaystyle f\left( Re^{it}\right)$ =  $\displaystyle e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi $.

So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi $.

Using Rouche's Theorem, which is a corollary of the Argument Principle.

$\displaystyle f( z) =\ z^{4} \ -\ 2z\ -\ 2$ in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$

$\displaystyle Let\ g( z) \ =\ -2$

For the smaller curve we have $\displaystyle z=\frac{1}{2} e^{i\theta }$,
$\displaystyle |f( z) -( -2) |\ \leq \ z^{4} \ +\ 2z\ \ =\ \frac{17}{16} \ < \ |g( z) |\ $

So, f(z) and g(z) have the same number of zeros in the smaller circle, which is 0.

$\displaystyle Now,\ let\ g( z) \ =\ z^{4}$
For the larger curve we have $\displaystyle z=\frac{3}{2} e^{i\theta }$
$\displaystyle |f( z) \ -\ z^{4} |\ \leq |2z\ +\ 2\ |\ \ =\ 5\ < \ |g( z) |\ =\ \frac{81}{16} \ \ $

So, f(z) and g(z) have the same number of zeros in the larger circle, which is 4.
Thus, all four zeros of f(z) are in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$

No problem! 

$\displaystyle f( z) \ =\ \frac{g( z)}{( z-z_{0})^{l}}$

where $\displaystyle g( z)$ is analytic in $\displaystyle |z\ -\ z_{0} |\ < \ R\ and\ g( z_{0}) \ \neq 0$
Then,
$\displaystyle \frac{f'( z)}{f( z)} \ =\ \frac{g'( z)}{g( z)} \ -\ \frac{l}{z\ -\ z_{0}}$

Since $\displaystyle g( z_{0}) \ \neq \ 0,\ then\ \frac{g'( z)}{g( z)}$ is analytic at $\displaystyle z_{0}$ , so the residue of that first term is 0. So for the residue of the whole thing, we read the coefficent of the second term which is -l. So we get

$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =-l$

By Cauchy's theorem, we get:
\begin{equation*}
0\ =\ \int _{\gamma } e^{iz^{2}} dz\ =\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} ie^{i\left( Re^{i\theta }\right)^{2}} Re^{i\theta } d\theta \ +\ \int ^{0}_{R} e^{ir^{2} \ e^{i\pi /2}} e^{\pi /4} \ dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{ir^{2}( cos\pi /2\ _{\ } +\ isin\pi /2)} dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
\\
\\
\end{equation*}

Now, just handling the middle integral, we get:


\begin{equation*}
\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \\
=R\int ^{\pi /4}_{0} e^{iR^{2} cos2\theta \ -\ R^{2} sin( 2\theta ) \ } ie^{i\theta } d\theta \\
This\ is\ estimated\ above\ by:\\
\\
R\int ^{\pi /4}_{0} e^{-R^{2} sin( 2\theta ) \ } d\theta \ \leq R\int ^{\pi /4}_{0} e^{-2R^{2}\frac{\theta }{\pi } \ } d\theta \ < \ \frac{\pi }{2R}\\
\end{equation*}


\begin{equation*}
So\ the\ limit\ as\ R\ \rightarrow \ \infty ,\ this\ will\ be\ 0,\ and\ we\ will\ get:\\
0=\int ^{R}_{0} e^{ix^{2}} dx\ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
0=\int ^{\infty }_{0}\left( cos\left( x^{2}\right) +\ isin\left( x^{2}\right) dx\right) \ -\ \frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\\
\frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{1+i}{\sqrt{2}} \ \frac{\sqrt{\pi }}{2} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{\sqrt{2\pi }}{4} \ +\ i\ \frac{\sqrt{2\pi }}{4} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\\
\end{equation*}

So, we separate imaginary and real parts and we get

\begin{equation*}
\int ^{\infty }_{0} cos\left( x^{2}\right) dx\ =\ \int ^{\infty }_{0} sin\left( x^{2}\right) dx\ =\frac{\sqrt{2\pi }}{4} \
\end{equation*}

\begin{equation*}
F( z) \ =\ \frac{-1}{z}\\
\\
The\ function\ is\ analytic\ in\ all\ of\ Re\ z\  >\ 0\\
\\
So\ we\ just\ want\ F( end\ point) \ -\ F( first\ point)\\
\\
F( 1-i) \ -\ F( 1\ +\ i)\\
\\
=\ \frac{-1}{1-i} \ +\ \frac{1}{1+i}\\
\\
=\ \frac{-1\ -\ i\ +\ 1\ -\ i\ }{( 1-i)( 1+i)} \ \\
\\
=\ \frac{-2i}{-2}\\
\\
=\ i
\end{equation*}

\begin{equation*}
\int _{|z|\ =\ 1}\frac{sin( z)}{z}\\
\\
=\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0}\\
\\
Set\ \zeta ( z) \ =\ sin( z)\\
\\
So,\ by\ Cauchy's\ formula,\\
\\
f( z) \ =\ \frac{1}{2\pi i}\int _{\gamma } \ \frac{\zeta ( z)}{\zeta \ -\ z}\\
\\
\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0} \ =\ ( 2\pi i) \zeta ( 0) \ \\
\\
=\ ( 2\pi i)( sin\ 0) \ =\ 0\ \ \\
\end{equation*}