# Thread: how to calculate residium at infinity?

1. ## how to calculate residium at infinity?

i have such a function
$
z^3 \sin \frac{1}{3}
$

i need to calculate its residium at z=infinity

if i substitue infinity instead of a"" into the formal formula
$res(f(x),a)=\lim_{x->a}(f(x)(x-a))$

i get infinity
am i correct?

2. Originally Posted by transgalactic
i have such a function
$
z^3 \sin \frac{1}{3}
$

i need to calculate its residium at z=infinity

if i substitue infinity instead of a"" into the formal formula
$res(f(x),a)=\lim_{x->a}(f(x)(x-a))$

i get infinity
am i correct?
The best way to find the residue at infinity is to replace "z" by "1/z" and find the residue at z=0.

Here, " $sin(\frac{1}{3})$" is a constant so replacing z by 1/z gives $\frac{sin(\frac{1}{3})}{z^3}$ which has a pole of order 3 at z=0. That has residue 0 at z= 0 so the residue of $sin(\frac{1}{3})z^3$ at infinity is also 0.

3. i am sorry i wrote the function in a wrong way
its
$
z^3 \sin \frac{1}{z}
$

$
\frac{1}{z^3} \sin z
$

so the third derivative of sine is -cos z
and divide by 3!
and put 0 instead of z
and in the end i get
-1/6
correct?

4. Originally Posted by transgalactic
i am sorry i wrote the function in a wrong way
its
$
z^3 \sin \frac{1}{z}
$

$
\frac{1}{z^3} \sin z
$

so the third derivative of sine is -cos z
and divide by 3!
and put 0 instead of z
and in the end i get
-1/6
correct?
Note that $\frac{\sin z}{z^3}$ has a double pole at z = 0.

But in fact it's trivial to get the residue just by writing out the first few terms of the Laurent series around z = 0. Then it's crystal clear that the residue is .....