I am tasked with finding whether the function $f(z) = \cos(z/(1-z))$ can be developed into a Laurent series around $z = 1$, and if so what is the radius of convergence and what is the residue?
So far what I got is:
- The function has an isolated singularity at $z = 1$ so indeed it can be developed into a Laurent series at that point.
- There are two different Laurent series expansions around $z = 1$, one for $|z| < 1$ (with radius of convergence 1 I suppose?) and one for $|z| > 1$ (with infinite radius of convergence). Is this right?
- If the above is correct, how would I go about developing the Laurent series from here for these two different domains? I know I can write $z/(1-z)$ as $-\frac{1}{z-1} - 1$ and can then sub that into the formula for $\cos(z)$ Taylor series, but not sure how that interacts with the different domains or how to simplify from there?
Don't overthink too much; you can write straightforwardly : $$ \begin{align} f(z) &= \cos\left(\frac{z}{z-1}\right) \\ &= \cos\left(1+\frac{1}{z-1}\right) \\ &= \cos(1)\cos\left(\frac{1}{z-1}\right) - \sin(1)\sin\left(\frac{1}{z-1}\right) \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} \frac{\cos(1)}{(z-1)^{2k}} - \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \frac{\sin(1)}{(z-1)^{2k+1}} \end{align} $$ where we used the Taylor series of the trigonometric functions. Alternatively, you may have used the identity $\cos t = \frac{e^{it}+e^{-it}}{2}$ and gather the terms of the exponential series together according to the powers of $i$. In consequence, the residue associated to this essential singularity turns out to be $\mathrm{Res}_{z=1} f(z) = -\sin(1)$. As for the radii of (the annulus of) convergence, they are given respectively by $r = \displaystyle \limsup_{k\to\infty} |a_{-k}|^{1/k} = 0$ and $\frac{1}{R} = \displaystyle \limsup_{k\to\infty} |a_k|^{1/k} = 0$, hence $R = \infty$.