Find the Taylor Series expansion of the given analytic function $f(z)$, centered at point $z_0$; find the disk of convergence.
a) $f(z)=\frac{1}{-2+3i-z}$ $z_0=3$
b) $f(z)=(2-z)\cos{(3z^2)}$ $z_o=0$
I know that I have to start off by finding the derivatives of each function until they start to repeat. From there I need to use the following formula to find the Taylor coefficients: $a_n=\frac{f^{(n)}(z_0)}{n!}$
For part (b) I want to confirm that I should be using the product rule to compute the derivatives.
For both part (a) and (b) I also get a little confused at the very end when it is time to actually write out the Taylor series expansion. How do I know where to find the disk of convergence?
Any help would be appreciated. Maybe if I can see a solution of just one of them I can continue with the other?
Best Answer
a) $f(z) = \frac{1}{-2+3i-z} = \frac{1}{-5+3i-(z-3)}$ converges in $|z-3| \lt |-5+3i|=\sqrt{34}$
also using geometric series expansion we have $$ f(z) = \frac{1}{-5+3i-(z-3)} = \frac{1}{-5+3i}\cdot\frac{1}{1-\frac{z-3}{-5+3i}} = \sum_{n=0}^{\infty}\frac{(z-3)^n}{(-5+3i)^{n+1}} $$
b) $f$ is entire and plugging $3z^2$ in the Taylor series of $\cos$ we have $$ f(z) = 2\cos(3z^2)-z\cos(3z^2) = \sum_{n=0}^{\infty}\frac{2\cdot 9^n\cdot(-1)^n}{(2n)!}z^{4n} - \sum_{n=0}^{\infty}\frac{9^n\cdot (-1)^n}{(2n)!}z^{4n+1} = $$
$$ = 2-z-9z^4+\frac{9}{2}z^5+\dots $$