I can't see how to express those in simpler form.

[Juggler123]

I need to find and classify the singular points and find the residue at each of these points for the following function;

f(z) = \frac{z^{1/2}}{z^{2}+1}

I can see that the singular points are at z=i and z=-i but have no idea how to classify them or find the residue at each point.

I know finding the residue depends on if the singular points are removable singulairties, poles or zeros of certain orders but don't know to classify z=i or z=-i.

Any help would be brillant, thankyou.

 

[HallsofIvy]

I need to find and classify the singular points and find the residue at each of these points for the following function;

f(z) = \frac{z^{1/2}}{z^{2}+1}

I can see that the singular points are at z=i and z=-i but have no idea how to classify them or find the residue at each point.

I know finding the residue depends on if the singular points are removable singulairties, poles or zeros of certain orders but don't know to classify z=i or z=-i.

Any help would be brillant, thankyou.

A function has a "pole of order n" at z= a if it can be written as a power series in z- a, the "Laurent series", including negative powers down to (z-a)^{-n}.
At an "essential singularity", such a power series includes all negative powers.
If there is a pole at z= a, the residue at z= a is the coefficient of (z-a)^n in
the powers series. We can write the example you give as \frac{z^{1/2}}{z+ i}\frac{1}{z- i}. Since \frac{z^{1/2}}{z+i} is analytic at z= i, it can be written as a power series in z- i with non-negative powers of z (Taylor's series)- which is of course, just it value at z= i. Multiplying that by (z-i)^{-1} gives a Laurent series down to (z- i)^{-1} so this function has a "pole of order 1" at z= i and its residue there is the same as the constant term of theTaylor's series for \frac{z^{1/2}}{z+ i} around z= i which, after multplying by \frac{1}{z- i}= (z- i)^{-1} gives the coefficient of (z- i)^{-1}.

In other words because z^{1/2} is analytic at z= i and z= -i, \frac{z^{1/2}}{z^2+ 1} has a pole of order 1 at both points. The residue at z= i is \frac{i^{1/2}}{i+ i}= \frac{i^{1/2}}{2i} and the residuce at z= -i is \frac{(-i)^{1/2}}{-i- i}= -\frac{(-i)^{1/2}}{2i}.