# Quiz 3

Compute
$\int_0^{2\pi} \frac{dt}{a+\cos t}$
assuming that a is a real number satisfying a>1.

Solution:
Let z=e^(it) so that cos t = (z+1/z)/2, dt = dz / (i z)

Then we have to compute the integral
$\int_{|z|=1} \frac{2dz}{i (z^2+2a z + 1)}$

The roots of the denominator are at the points
$-a\pm \sqrt{a^2-1}$

Since a>1, both these numbers are real. The one with the minus sign is outside the unit circle, while the one with plus sign is inside:
$\sqrt{a^2-1}-a = \frac{-1}{\sqrt{a^2-1}+a}$
and this number is clearly between -1 and 0.

Thus the integral is equal to 2 pi i times the residue of the integrand at
$z=\sqrt{a^2-1}-a$
which is equal to
$\frac{2}{i(\sqrt{a^2-1}-a - (-\sqrt{a^2-1}-a))}=\frac{1}{i\sqrt{a^2-1}}$
Hence the integral is equal to
$\frac{2\pi}{\sqrt{a^2-1}}$

Find the coefficients in the power series expansion
$\frac{1+z-i}{1+z+i}=\sum a_n (z+1)^n$
For what values of z is the expansion valid?

Let w=z+1. Then we want to expand (w-i)/(w+i) in powers of w. Rewrite it as 1 - 2i/(w+i), or as 1 - 2/(1-iw). Finally using geometric series we see that 1 - 2/(1-iw)= 1-2(1+ iw + (iw)^2 + ...), with the expansion valid for |iw|<1.

Thus
$a_n= -2 i^n$
for n>0 and a_0 = -1.

The expansion is valid for |z+1|<1.

Suppose $f$ is analytic on the whole complex plane and $f(1/n)=1/n^2$ for all natural numbers $n$.
(a) Show that $f(0)=0$.
(b) Show that $f$ has a zero of order $2$ at $z=0$.

Solution:
(a) Since f is analytic, it is in particular continuous at point 0, hence
$f(0)=\lim_{n \to \infty} f(1/n) = 0$
(b) Solution 1: Let g(z)=f(z)-z^2. The function g is analytic, g(0)=0, g(1/n)=0. Hence g has a non-isolated zero at z=0. Since non-zero analytic functions have only isolated zeroes, g must be identically zero. Hence f(z)=z^2 for all z, and hence it has a zero of order 2 at z=0.

Solution 2: f(0)=0. Suppose that f has a zero of order k at z=0. Then for small enough z, f(z) is approximately equal to c z^k for some non-zero constant c (in the sense that
$\lim\limits_{z\to 0} \frac{f(z)}{z^k}=c$
)
In particular
$\lim\limits_{n\to \infty} \frac{f(1/n)}{(1/n)^k}=c$
On the other hand this is the limit of n^(k-2) as n tends to infinity. It is a non-zero number if and only if k=2.