Hi, how I can find Laurent series of this function at the point $z = -1$.

Question

$$f(z) = z^3\cdot \cosh\Bigl(\frac{1}{z+1}\Bigr)$$

I try this: $$\cosh \Bigl(\frac{1}{z+1}\Bigr) = z^3\cdot\sum_{n=0}^{\infty}\frac{1}{(2n!)(z+1)^{2n}}$$ But I do not understand how I can expand in powers of $z + 1$.

Answer 1

$$ z^3 = (z+1-1)^3 = (z+1)^3 -3(z+1)^2 + 3(z+1) - 1 \text{.} $$

Then reindex your sums by power of $z+1$. Some sums will start before other sums. Pop off those terms -- so that you have a few explicit terms plus sums that all have the same range of index and the same power of $z+1$. Now combine the sums.