# Laurent series , function representation

Write the Laurent series around zero for the entire function $f(z)=z^2e^{3z}$

I'm a little confused on how to represent the complex functions by series, as I did in the calculation of real functions, but do not know if it's right

$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\rightarrow e^{3z}=\sum_{n=0}^\infty \frac{3^nz^n}{n!}\rightarrow z^2e^{3z}=\sum_{n=0}^\infty \frac{3^nz^{n+2}}{n!}$$

ii) Find the Laurent series representation for $f(z)=z^2\sin(\frac{1}{z^2})$ where $0<|z|<\infty$

• should be $\sum _{n=0}^{\infty } \frac{3^nz^{n+2}}{n!}$ – Lozenges May 5 '15 at 18:43
• You may write $$z^2e^{3z}=\sum_{n=2}^{\infty} \frac{3^{n-2}z^{n}}{(n-2)!}$$ – Olivier Oloa May 5 '15 at 18:44
• @OlivierOloa But my answer is correct? The procedure to find the representation of a complex function is the same as real functions? – Roland May 5 '15 at 18:48
• @askazy Your procedure and your result are correct. – Olivier Oloa May 5 '15 at 18:55
• @OlivierOloa My biggest question is when there is a restriction on z, take a look at what added kindly. – Roland May 5 '15 at 19:01

## 1 Answer

For the second part, you may write $$\sin\left(\frac{1}{z^2}\right)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!\:z^{2(2n+1)}},\quad z \neq0,$$ giving $$f(z)=z^2\sin\left(\frac{1}{z^2}\right)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!\:z^{4n}},\quad z \neq0.$$ since $\displaystyle u \longmapsto \sin(u)$ has a power series with an infinite radius of convergence.

• @askazy My bad, I'm going to correct it! Thanks. – Olivier Oloa May 5 '15 at 19:53
• @askazy This time it's OK. – Olivier Oloa May 5 '15 at 20:00
• @askazy Thanks! Let's delete our comments. – Olivier Oloa May 5 '15 at 20:03