Largest number for which a laurent series converges

Question

For part $(a)$ I got summation from $\sum^{\infty}_{n=0}(-1)^n\frac{z^{2n}}{(2n+1)!}$

Is this correct?

Could someone explain how to do part (b) because I have no idea where to start

Thanks

Answer 1

$$\sin 2z=\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{(2n+1)!}\implies\frac{\sin2z}z=\sum_{n=0}^\infty (-1)^n\frac{z^{2n}}{(2n+1)!}$$

For part (b):

$$\frac1{z-1}+\frac2{1-i+(z-1)}=\frac1{z-1}+\frac1{1-i}\frac1{1+\left(\frac{z-1}{1-i}\right)}$$

Thus, for

$$\left|\frac{z-1}{1-i}\right|<1\iff|z-1|<|1-i|=\sqrt2$$

we get:

$$\frac1{z-1}+\frac1{z-i}=\frac1{z-1}+\frac1{1-i}\left(1-\frac{z-1}{1-i}+\left(\frac{z-1}{1-i}\right)^2-\ldots\right)=$$

$$=\frac1{z-1}+\frac1{1-i}-\frac{(z-1)}{(1-i)^2}+\ldots$$