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Wednesday, April 11, 2012

Complex Analysis Part 2

Liouville's Theorem 
If a function $f(z)$ is analytic for all finite values of z, and is bounded then it is a constant.
Note:- $e^{z+2\pi i} = e^z$
Taylor's Theorem
If a function $f(z)$ is analytic at all points inside a circle C, with its centre at point a and radius R then at each point z inside C
Taylor's theorem is applicable when function is analytic at all points inside a circle.
 Laurent Series
If $f(z)$ is analytic on $C_{1}$ and $C_{2}$ and in the annular region R bounded by the two concentric circles $C_{1}$ and $C_{2}$ of radii $r_{1}$and $r_{2}$ ($r_{1} > r_{2}$) with their centre at a then for all z inside R
$a_{n}=\frac{1}{2\pi i}\int_{C_{1}}\frac{f(w)dw}{(w-a)^{(n+1)}}$
$b_{n}=\frac{1}{2\pi i}\int_{C_{1}}\frac{f(w)dw}{(w-a)^{(-n+1)}}$
Singular points
If a function $f(z)$ is not analytic at point z=a then z=a is known as a singular point or there is a singularity of $f(z)$ at z=a for example
z=2 is a singularity of $f(z)$
Pole of order m
If $f(z)$ has singularity at z=a then from laurent series expansion
and we say that function $f(z)$ is having a pole of order m at z=a. If m=1 then point z=a is a simple pole.
The constant $b_{1}$ , the coefficent of $(z-z_{0})^{-1}$ , in the Laurent series expansion is called the residue of $f(z)$ at singularity $z=z_{0}$
$b_{1}=Res_{z=z_{0}}f(z) = \frac{1}{2\pi i}\int_{C_{1}}f(z)dz$
Methods of finding residues
1.  Residue at a simple pole
if $f(z)$ has a simple pole at z=a then
$Res f(a) =\lim_{z\rightarrow a}(z-a)f(z) $
2. If $f(z)=\frac{\Phi(z)}{\Psi (z)}$
and $\Psi(a)=0$ then
$Res f(a)=\frac{\Phi(z)}{\Psi^{'} (z)}$
3. Residue at pole of order m
If $f(z)$ is a pole of order m at z=a then
$Res f(a)= \frac{1}{(m-1)!}\left \{ \frac{d^{m-1}}{dz^{m-1}}(z-a)^{m}f(z) \right \}_{z=a}$
 Residue Theorem
If $f(z)$ is analytic in closed contour C excapt at finite number of points (poles) within C, then
$\int_{C}f(z)dz = 2\pi i \textsl { [sum of the residues at poles within C]}$

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