$±i$ in residue calculations

Question

So we need to find the singularities and thus the residues of the function $z/2z^4+5z^2+2$. Since $2z^4+5z^2+2=(2z^2+1)(z^2+2)$, this function clearly has a simple pole at each of the points $z=\pm i\sqrt{2}$ and $z=\pm i/\sqrt{2}$. Only the latter two singularities lie inside the contour so we need only consider the residues at these points. For the residues we have that, using the results of Lemma 4.6.9, $$\operatorname*{res}\left(\frac{z}{2z^4+5z^2+2},\pm \frac{i}{\sqrt{2}}\right)=\frac{z}{8z^3+10z}\Biggr\rvert_{z=\pm i/\sqrt{2}}=\frac{1}{6}$$

This is an example of calculating residue, what I don't understand from all the examples is how do they sub in the values of $z$ given that its $±$ values, which one to use and how the get the answers. I have tried many attempts at subbing the value in but I don't get the desired result. Could someone explain the concept of $±i$ values and how to work residue in a step by step procedure?

Answer 1

A solution following the quoted results of the OP: Using the formula for residues of quotients (see the wikipedia article):

If $f(z)=\frac{p(z)}{q(z)}$ with $p(c)\not=0$, $q(c)=0$, and $q'(c)\not=0$, then the residue at $z=c$ is given by $$ \frac{p(c)}{q'(c)}. $$

In this case, $p(z)=z$ and $q(z)=2z^4+5z^2+2$. Therefore, $q'(z)=8z^3+10z$. Hence, the residue is given by substitution of the poles (roots of $q(z)$) into this fraction, which is exactly $$ \frac{z}{8z^3+10z} $$ as given by the OP. Then, substituting one of the given values, say $\frac{i}{\sqrt{2}}$ results in $$ \frac{\frac{i}{\sqrt{2}}}{8\left(\frac{i}{\sqrt{2}}\right)^3+10\left(\frac{i}{\sqrt{2}}\right)}=\frac{\frac{i}{\sqrt{2}}}{-\frac{4i}{\sqrt{2}}+\frac{10i}{\sqrt{2}}}= \frac{\frac{i}{\sqrt{2}}}{8\left(\frac{i}{\sqrt{2}}\right)^3+10\left(\frac{i}{\sqrt{2}}\right)}=\frac{\frac{i}{\sqrt{2}}}{\frac{6i}{\sqrt{2}}}. $$ Clearing fractions, you get $\frac{1}{6}$. You will get a similar result if you substitute $-\frac{i}{\sqrt{2}}$

Answer 2

One way to get this is as follows (I am not sure where they get $8z^3+10z$ from, perhaps you could include Lemma 4.6.9):

Let's begin by factoring the denominator: $$ 2z^4+5z^2+2=(2z^2+1)(z^2+2)=2\left(z+\frac{i}{\sqrt{2}}\right)\left(z-\frac{i}{\sqrt{2}}\right)(z+i\sqrt{2})(z-i\sqrt{2}) $$

Therefore, to get the residue at $z=\frac{i}{\sqrt{2}}$, we do the following: $$ \frac{z}{2z^4+5z^2+2}=\frac{1}{z-\frac{i}{\sqrt{2}}}\cdot\frac{z}{2\left(z+\frac{i}{\sqrt{2}}\right)(z+i\sqrt{2})(z-i\sqrt{2})} $$

Observe that the factor on the right is defined near $z=\frac{i}{\sqrt{2}}$, so the residue is the value of this function: $$ \frac{\frac{i}{\sqrt{2}}}{2\left(\frac{i}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{i}{\sqrt{2}}+i\sqrt{2}\right)\left(\frac{i}{\sqrt{2}}-i\sqrt{2}\right)} $$ Simplifying, we get that this simplifies to $$ \frac{i}{4i\left(-\frac{1}{2}+2\right)}=\frac{1}{6} $$

Answer 3

When calculating residues, only one value of $z$ should be examined at a time.

The author is just being lazy and really means to say this:

$$\operatorname{res}\left(\frac{z}{2z^4+5z^2+2},\frac{i}{\sqrt{2}}\right)=\frac16$$ and $$\operatorname{res}\left(\frac{z}{2z^4+5z^2+2},-\frac{i}{\sqrt{2}}\right)=\frac16$$