- #1
jrp131191
- 18
- 0
Hi, I am still trying to learn latex so this might fail but anyway, I am trying to establish the identity
[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?
[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?