write Taylor series for the function f(z)=1/z centered at z=i

Question

I know that I have to calculate $a_n$ first. $a_n=\frac{1}{n!} f^{(n)}(z_0)$

so I arrived at this formula for $f^{(n)}(z)=(-1)^n.\frac{n!}{z^{n+1}}$

then we have $a_n=\frac{(-1)^n}{i^{n+1}}$

so from Taylor series we have $$f(z)=\sum_{n=1}^{\infty} a_n (z-z_0)^n $$

so for f(z)=1/z we have $$f(z)=\frac{(-1)^n}{i^{n+1}} .(z-i)^n$$ but I can't confidentially say that this approach is true.If you see any problems with my solution please tell me.

Answer 1

You can either say.

$\frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$

evaluated at $z=i$

$a_n = (-1)^n (i)^{-n-1}\\ -1 = i^2\\ a_n = (i)^{n-1}\\ f(z) = \sum_\limits{0}^{\infty} i^{n-1} {(z-i)^n}$

Alternatively, you could use the formula for a geometric series.

$\sum_\limits{n=0}^\infty z^n = \frac {1}{1-z}$ and manipulate your function to fit this form.

$f(z) = \frac {1}{z} = \frac {1}{i+(z-i)} = \frac {i}{-1+(z+i)i}= -i\sum i^n(z-i)^n = \sum i^{n-1}(z-i)^n$