# Laurent series of $g(z) =( e^{3iz} - 3 e^{iz} +2)/z^3$ around $0$

I am supposed to find the Laurent expansion of

$$g(z) = f(z)/z^3 = (e^{3iz} - 3 e^{iz} +2)/z^3$$

Now using Taylor I get that

$$f(z) = \sum_{n=0}^\infty \frac{(3iz)^n}{n!} - 3\frac{(iz)^n}{n!} + 2 = \sum_{n=0}^\infty \frac{(3iz)^n}{n!} - 3\frac{(iz)^n}{n!} + \left( \frac{1}{2}\right)^n$$

Now when $|z|<1$ we can write the Laurent series as

$$g(z) = \overbrace{\sum_{n=0}^{\infty} (1-z^3)^n}^{\Large z^3} \left( e^{3iz} - 3 e^{iz} +2\right)$$

But I do not know how to continue, the book gives the answer as

$$g(z) = \sum_{n=2}^\infty \frac{3(3^{n-1}-1)}{n!}i^n z^n = \sum_{n=-1}^\infty \frac{3(3^{n+2}-1)}{(n+3)!}i^{n+3}z^n \quad 0 < |z| < \infty$$

but I do not quite see how to obtain this.

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I can't understand how the answer you say your book (which one, by the way?) gives can be correct, as the sum given by them "begins" with $\,z^{-1}\,$ and not with $\,z^{-3}\,$ ... – DonAntonio Nov 25 '12 at 11:51

$$f(z)=\sum_{n=0}^\infty \frac{(3iz)^n}{n!} - \sum_{n=0}^\infty3\frac{(iz)^n}{n!} + 2=\sum_{n=0}^\infty\frac{(3^n-3)i^nz^n}{n!}+2\;\;\;(**)$$
But it's easy to check that the above expression is zero for $\,n=0\,$ , so:
$$(**)\;=\sum_{n=1}^\infty\frac{3(3^{n-1}-1)}{n!}i^nz^n+2\Longrightarrow$$
$$\Longrightarrow g(z)=\frac{2}{z^3}+\sum_{n=1}^\infty\frac{3(3^{n-1}-1)i^n}{n!}\,z^{n-3}$$