\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\p}{\partial}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Find the real and imaginary parts of the functions $\cos z$ and
$\sin z$ where $z=x+iy$.  Verify the Cauchy-Riemann equations in
each case and show that $\ds\frac{d}{dz}(\sin z)=\cos z$.


\item[b)]
Prove that, unless $z$ is a real number

$$|\sin z|^2+|\cos z|^2>1.$$


\item[c)]
Evaluate the integral $\ds\int_C\tan zdz$ where $C$ is the
straight line segment from $z=0$ to $z=\frac{\pi}{2}(1+i)$.

Express your answer in the form $a+ib$ where $a$ and $b$ are real.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\cos z=\cos(x+iy)=\cos x\cos iy-\sin x\sin iy$

$\hspace{0.5in}=\cos x\cosh y-i\sin x\sinh y$

Similarly $\sin z=\sin x\cosh y+i\cos x\sinh y$

For $\cos z$

$\begin{array}{ll} \ds \frac{\p u}{\p x}=-\sin x\cosh y & \ds
\frac{\p v}{\p y}=-\sin x\cosh y\\ \ds \frac{\p u}{\p y}=\cos
x\sinh y & \ds -\frac{\p v}{\p x}=\cos x\sinh y\end{array}$

For $\sin z$

$\begin{array}{lr} \ds \frac{\p u}{\p x}=\cos x\cosh y & \ds
\frac{\p v}{\p y}=\cos x\cosh y\\ \ds \frac{\p u}{\p y}=\sin
x\sinh y & \ds -\frac{\p v}{\p x}=\sin x\sinh y\end{array}$

so the Cauchy-Riemann equations are satisfied, and the partial
derivatives are all continuous.

$\ds\frac{d}{dz}(\sin z)=\frac{\p u}{\p x}+i\frac{\p v}{\p x}=\cos
x\cosh y-i\sin x\sinh y=\cos z$


\item[b)]
$|\sin z|^2+|\cos
z|^2=\sin^2x\cosh^2y+\cos^2x\sinh^2y$

$\hspace{0.7in}+\cos^2x\cosh^2y+\sin^2x\sinh^2y$

$=\cosh^2y+\sinh^2y=\cosh2y\geq1$ with equality iff $y=0$.


\item[c)]
$C: \, z=(1+i)t \hspace{0.2in} 0\leq t\leq\frac{\pi}{2}$

$\int_C\tan z=\int_0^{\frac{\pi}{2}}\tan(1+i)t(1+i)dt=
[-\log\cos(1+i)t]_0^{\frac{\pi}{2}}$

$\ds=-\log\cos\left(\frac{\pi}{2}+i\frac{\pi}{2}\right)=
-\log\left(-sin\left(i\frac{\pi}{2}\right)\right)$

$\ds=-\log\left(-i\sinh\frac{\pi}{2}\right)=
-\ln\sinh\frac{\pi}{2}+i\frac{\pi}{2}$

\end{itemize}

\end{document}