- #1

jrp131191

- 18

- 0

[itex]arctan(z) = \frac{1}{2i}log(\frac{1+iz}{1-iz})[/itex]

By using the power series of

[itex]log(1+iz)[/itex] and [itex]log(1-iz)[/itex]

I have found the power series for both of these functions as well as arctan(z)

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

[itex]log(1+iz) = -\Big( \sum_{n=0}^\infty\frac{(-1)^(n+1)((iz)^(n+1))}{n_1} \Big) [/itex]

[itex]log(1-iz) = -\Big( \sum_{n=0}^\infty\frac{(-1)^(n+1)((-iz)^(n+1))}{n_1} \Big) [/itex]

I then computed

[itex]\frac{1}{2i}(log(1+iz)-log(1-iz))[/itex]

I get a sum which doesn't really look anything like that expected. I have tried simplifying but still can't deduce that they are equal. I thought I could perhaps differentiate both sums n times and show that these are equal but that also gets very very messy.. can somebody point me in the right direction?