assuming that a is a real number satisfying a>1.
Let z=e^(it) so that cos t = (z+1/z)/2, dt = dz / (i z)
Then we have to compute the integral
The roots of the denominator are at the points
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:
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
which is equal to
Hence the integral is equal to
Find the coefficients in the power series expansion
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.
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 $z=0$.
(a) Since f is analytic, it is in particular continuous at point 0, hence
(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
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.
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