Question

show that when 0<|z-1|<2:
z/((z-1)*(z-3))=-3*(summation from n=0 to infinity of) ((z-1)^n/2^(n+2))-1/(2*(z-1))

Answer 1

\[\sum_{n=0}^{\infty} \frac{ (z-1)^{n} }{2^{n+2} } - \frac{ 1 }{ (2(n-1)) }\]

Answer 2

It's the Taylor series about \(z=1\) extended to include a single negative power of \(z\) term.
http://www.wolframalpha.com/input/?i=power+series+of+z%2F%28%28z-1%29%28z-3%29%29+about+z%3D1

Answer 3

Consider:$$\frac{z}{z-3}=\frac{z-3+3}{z-3}=1+\frac3{z-3}$$First we find the Taylor series of \(\frac1{z-3}\), which is trivial:$$\frac1{z-3}=-\sum_{n=0}^\infty\frac1{2^{n+1}}(z-1)^n\\\frac3{z-3}=-3\sum_{n=0}^\infty\frac1{2^{n+1}}(z-1)^n\\1+\frac3{z-3}=1-3\sum_{n=0}^{\infty}\frac1{2^{n+1}}(z-1)^n\\\frac{z}{z-3}=1-3\sum_{n=0}^{\infty}\frac1{2^{n+1}}(z-1)^n\\\frac{z}{(z-1)(z-3)}=\frac1{z-1}-3\sum_{n=0}^{\infty}\frac1{2^{n+1}}(z-1)^{n-1}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac1{z-1}-\frac3{2^0}(z-1)^{-1}-3\sum_{n=0}^\infty\frac1{2^{n+2}}(z-1)^n\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac2{2(z-1)}-\frac3{2(z-1)}-3\sum_{n=0}^\infty\frac1{2^{n+1}}(z-1)^n\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac1{2(z-1)}-3\sum_{n=0}^\infty\frac1{2^{n+1}}(z-1)^n$$

Answer 4

... and \(0\lt|z-1|\lt2\) so (1) we don't divide by \(0\) at any term in our expansion since \(z=1\) is a singular point and (2) note that \(|z-1|<2\implies|z-3|\lt0\), again because there lies a singular point of our original function.

Answer 5

oops those last two lines are supposed to have \(2^{n+2}\) not \(2^{n+1}\)

Answer 6

ALSO! for \(|z-3|\gt0\) we have obvious divergence, while for \(z=3\) (despite having a singularity in our original function) we end up with a divergent series:$$\sum_{n=0}^\infty\frac{(3-1)^n}{2^{n+1}}=\frac12\sum_{n=0}^\infty\frac{2^n}{2^n}=\frac12\sum_{n=0}^\infty1$$

Answer 7

I solved this already.