# 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.

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

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

this gives us a power series around $z_0 =-1+i$ that can be expressed as

$\sum_{n=0}^\infty \left(\frac 1 {2-i}\right)^{n+1}(z-z_0)^n$

• Thank you this really helped! Feb 12, 2014 at 18:59

We have that \begin{align} (1-z)^{-1} &= \big((2-i)-(z+1-i)\big)^{-1}=\frac{1}{2-i}\left(1-\frac{z+1-i}{2-i}\right)^{-1} \\ &=\frac{1}{2-i}\sum_{n=0}^\infty \left(\frac{z+1-i}{2-i}\right)^n\\&=\sum_{n=0}^\infty (2-i)^{-n-1}(z+1-i)^n. \end{align}

Note that the radius of convergence of this series is $r=|2-i|=\sqrt{5}$.

• Greatly appreciated thank you! Feb 12, 2014 at 18:59