# Residue of the function $z^3 cos \frac{1}{z-2}$

I'm trying to find the residue of the function $f(z) = z^3 cos \frac{1}{z-2}$

And I think the pole of this function is at z=2

From here I don't seem to have any ideas on proceeding further. If I'm not wrong, we have to expand cos??

Thanks for the help.

• I very much suspect that $z=2$ is an essential singularity. Commented Jan 12, 2016 at 22:56
• What I believe that Arthur is indicating is that pole should be replaced by singularity.
– robjohn
Commented Jan 12, 2016 at 23:42

\begin{align} &z^3=\{(z-2)+2\}^3=(z-2)^3+6(z-2)^2+12(z-2)+8\\ &\cos\left({1\over z-2}\right)=\sum_{n=0}^\infty{(-1)^n\over(2n)!(z-2)^{2n}}\\ &\therefore f(z)=\sum_{n=-\infty}^\infty a_n(z-2)^n\implies a_{-1}={1\over4!}-{12\over2!}=-{143\over24} \end{align}

• (+1) all three almost simultaneous answers are good (yours was 20 seconds earlier than mine).
– robjohn
Commented Jan 12, 2016 at 23:39
• @robjohn I deleted my answer when I saw our answers didn't match. After all, you're the great robjohn :) Commented Jan 12, 2016 at 23:48
• I first read the question as $z^2\cos\left(\frac1{z-2}\right)$, so my initial answer was for that. Then I noticed that it was $z^3$, not $z^2$.
– robjohn
Commented Jan 12, 2016 at 23:51

Substituting $z\mapsto z+2$ does not change the residue $$(z+2)^3\cos\left(\frac1z\right)=\left(z^3+6z^2+12z+8\right)\left(1-\frac1{2z^2}+\frac1{24z^4}+O\left(\frac1{z^6}\right)\right)$$ The coefficient of $\frac1z$ is $-6+\frac1{24}=-\frac{143}{24}$, which would be the residue.

• I know it's super late but could you describe why substituting $z\mapsto z+2$ does not change the residue? @robjohn Commented Mar 31, 2021 at 8:24
• The residue of $\sum\limits_{k=-\infty}^\infty a_k(z-b)^k$ at $z=b$ is $a_{-1}$. Substituting $z\mapsto z+b$ gives $\sum\limits_{k=-\infty}^\infty a_kz^k$ whose residue at $z=0$ is $a_{-1}$.
– robjohn
Commented Mar 31, 2021 at 8:42
• When expanding the function $f(z)$ at $z=a$ we write it as a series in powers of $z-a$ and the coefficient of $\frac1{z-a}$ is the residue at $a$.
– robjohn
Commented Mar 31, 2021 at 8:51
• Aha!! Thanks, @robjohn. Your comment clear my confusion. Commented Mar 31, 2021 at 9:31

Consider instead $$g(z)=(z+2)^3\cos\frac{1}{z}= (z+2)^3\sum_{n=0}^{\infty}(-1)^n\frac{1}{z^{2n}\,(2n)!}$$ which has an essential singularity at $0$. Since $$(z+2)^3=z^3+6z^2+12z+8$$ it shouldn't be difficult to find what terms have $z^{-1}$ when the product is expanded.