How to calculate tan^(-1) (2i) ??

Question

Hi everyone. I've got this question here

tan^(-1) (2i)

And also the answer, which is

(n + 1/2)(\pi) + i/2 * ln(3) for n(=0, \pm 1, \pm 2 ...)

Now I've got the identity for tan^(-1) z

Which is

tan^(-1) z = i/2 * log (i+z)/(i-z)

Now when z = 2i then I just sub in right? well obviously not because i'm asking this q here

I got tan^(-1) (2i) = i/2 log(-3) which i know is not defined.

So how does the answer have the period (n+1/2)pi and where does the i/2 go and how do ijust turn the -3 into a 3

Sorry about all the questions... have to be able to do these types of q's for an exam next month.

Thanks guys
Laura

Answer 1

ln(-1) = iπ

because: e^(ix) = cos(x) + isin(x), so if x = π

e^(iπ) = -1 --> iπ = ln(-1)

but actually ANY odd integer multiple works so write it as:
(it has to be odd so that cos is always -1

e^(i(2n +1)π) = -1 for n = 0 ± 1 ± 2 ...

so we can write it as:

ln(-1) = (2n +1)iπ for n = 0 ± 1 ± 2 ...

so first write:

i/2 ln(-3) = i/2 ln([-1]*[3]) = i/2[ln(3) + ln(-1)]

= i/2 ln(3) + i/2ln(-1)
= i/2 ln(3) + i/2(2n +1)iπ for n = 0 ± 1 ± 2 ...
= i/2 ln(3) - π/2(2n +1) for n = 0 ± 1 ± 2 ...