Question

If f(z)=(z+1)/(z-1), then find f^1991(2+i)

Answer 1

Full solution will get a medal (please tell me how-first time user)

Answer 2

I got (2-i)^1991 and then what?

Answer 3

\(\large\tt \color{black}{ f^1991(2+i)=(\dfrac{2+i+1}{2+i-1})^{1991}}\)

\(\large\tt \color{black}{ =(\dfrac{3+i}{1+i})^{1991}}\)

\(\large\tt \color{black}{ =(\dfrac{3+i}{1+i}\times \dfrac{1-i}{1-i})^{1991}}\)

\(\large\tt \color{black}{=(2-i)^{1991}}\)

\(\large\tt \color{black}{=(2-i)\times(2-i)^{1990}}\)

\(\large\tt \color{black}{=(2-i)\times(2-i)^{2(945)}}\)

\(\large\tt \color{black}{=(2-i)\times(2^2-2i+i^2)^{945}}\)


\(\large\tt \color{black}{=(2-i)\times(3-i)^{2(945)}}\)

\(\large\tt \color{black}{=(2-i)(3-i)(3-1)^{2(472)}}\)


like this u have to solve it untill u get the answer

Answer 4

second last term is incorrect ,it should be this
\(\large\tt \color{black}{=(2-i)\times(3-i)^{945}}\)

Answer 5

but then it would go on infinitely

Answer 6

yes i think so