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Power series expansion of $$f(z)=\frac{1}{3-z}$$ around $4i$.

I calculated the radius of convergence to be $5$, and I obtained the power series:

$$\sum \limits_{n=0}^\infty \frac{n!(z-4i)^n}{(3-z)^{-(n+1)}}$$

The answer key has the same, except they don't have my $n!$ term. Why? What have I done wrong?

$$f=\frac{1}{3-z}$$ $$f^{(1)}=(3-z)^{-2}$$ $$f^{(2)}=2(3-z)^{-3}$$ $$f^{(3)}=6(3-z)^{-4}$$ $$\vdots$$ $$f^{(n)}=n!(3-z)^{-(n+1)}$$ It would seem.

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The coefficient of $(x-a)^n$ in the Taylor series of $f(x)$ around $x=a$ is $f^{(n)}(a)/n!$. You forgot to divide by $n!$.

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  • $\begingroup$ Yes, true, thank you, I'm silly! $\endgroup$ Jun 13, 2015 at 2:51

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