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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 $0<|z|<\infty$

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  • $\begingroup$ should be $\sum _{n=0}^{\infty } \frac{3^nz^{n+2}}{n!}$ $\endgroup$
    – Lozenges
    May 5, 2015 at 18:43
  • $\begingroup$ You may write $$z^2e^{3z}=\sum_{n=2}^{\infty} \frac{3^{n-2}z^{n}}{(n-2)!}$$ $\endgroup$ May 5, 2015 at 18:44
  • $\begingroup$ @OlivierOloa But my answer is correct? The procedure to find the representation of a complex function is the same as real functions? $\endgroup$
    – Roland
    May 5, 2015 at 18:48
  • $\begingroup$ @askazy Your procedure and your result are correct. $\endgroup$ May 5, 2015 at 18:55
  • $\begingroup$ @OlivierOloa My biggest question is when there is a restriction on z, take a look at what added kindly. $\endgroup$
    – Roland
    May 5, 2015 at 19:01

1 Answer 1

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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.

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