# Express $(1-z)^{-1}$ as a power series around $z_0=-1+i$.

I need to express $(1-z)^{-1}$ as a power series in powers of $(z+1-i)$.

I would like some guidance on the complex analogue of power series and in writing out this particular case.

Many thanks for any answers!

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We see that $(1-z)$ can be expressed as $(1-z + 1 - 1 + i-i)= ((2-i)-(z+1-i))$

We want to get back to the form $(1-a)^{-1}$ to equate to $\sum_{n=0}^\infty a^n$ so we divide by $(2-i)$ giving us $\frac 1 {2-i}\left(1-\frac{z+1-i}{2-i}\right)^{-1}$

We then plug the second part into our formula getting $\left(\frac 1 {2-i}\right)\sum_{n=0}^\infty \left(\frac{z+1-i}{2-i}\right)^n$

Then we bring the $\left( \frac 1 {2-i} \right)^{n}$ out and combine with our outside term getting

$\sum_{n=0}^\infty \left(\frac 1 {2-i}\right)^{n+1}(z+1-i)^n$

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Thank you this really helped! –  AnalysisisKey Feb 12 at 18:59
Note that the radius of convergence of this series is $r=|2-i|=\sqrt{5}$.