A couple of GRE questions Look for help with the following GRE questions
Question 1. If $C$ is the circle in the complex plane whose equation $|z|=\pi$, oriented counterclockwise, find the value of the integral $\oint_C(\cos z-z\cos\frac{1}{z})dz$.
Question 2. How to show the sequence $\{x_n\}_{n=1}^\infty$ definted by $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$, $x_1\ne 0$ converge. 
Question 3. Let $L$ be the curve whose equation in the polar coordinates $r$ and $\theta$ is $r^2=4\cos 2\theta$. Fine the largest value of $y$ such that the point with rectangular coordinates $(x,y)$ is on $L$.
Question 4. Consider the set $S$ of all real-valued functions defined on $[a, b]$, $a<b$. Is it true that if the inverse of $f$ is a constant function, then $f$ is a constant function?

For the first, $\oint_C \cos z dz = 0$ since $\cos$ is an entire function. Thus, only the term $z\cos\frac{1}{z}$ is relevant. That term has a single pole at $z=0$, and by the residue theorem you therefore have $$
  \oint_C z\cos\frac{1}{z} dz = i2\pi R
$$
where $R$ is the residue (i.e. the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$) at $0$. Now, $\cos z = \frac{1}{2}(e^{iz} + e^{-iz})$, hence the coefficient of $z^2$ in the taylor expansion of $\cos z$ is $\frac{1}{2}(\frac{-1}{2} + \frac{-1}{2})$ = $-\frac{1}{2}$. Which is the same as the coefficient of $z^{-2}$ in the laurent expansion of $\cos\frac{1}{z}$, and hence the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$, i.e. the residue. Together, you get that $$
  \oint_C \cos z - z\cos\frac{1}{z} dz =  0 - i2\pi\frac{-1}{2} = i\pi
$$

For the second, first find a fixed point, i.e. solve $x_{n+1} = x_n$. You get $$
   \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) = x_n \iff
   \frac{1}{x_n} = \frac{x_n}{2} \iff
   x_n = \sqrt{2}
$$
Then show that $\sqrt{2} < x_{n+1} < x_n$ if $x_n > \sqrt{2}$, which works similarly $$
\begin{eqnarray}
   \sqrt{2} < \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) < x_n &\iff&
   \sqrt{2} - \frac{x_n}{2} < \frac{1}{x_n} < \frac{x_n}{2} \iff \\
   2\sqrt{2}x_n - x_n^2 < 2 < x_n^2 &\iff&
   2 - (x_n - \sqrt{2})^2 < 2 < x_n^2
\end{eqnarray}
$$
The last inequality shows not only that the sequence decrases monotonically if started at any value larger than $\sqrt{2}$, it also shows that regardless of where it it started, it always immediatly assumes a value larger than $\sqrt{2}$. Thus, the sequence converges regardless of its initial value (Strictly speaking, the above verified that only for positive $x_1$, but things work identically for negative $x_1$)
