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

## Homework Equations

sin^(-1)(z) = -i * log[iz + (1-z^2)^(1/2)]

cos^(-1)(z) = -i * log[z + i(1-z^2)^(1/2)]

## The Attempt at a Solution

I've tried a number of attemps.

I first tried the most obvious and I did sin^(-1)(z) / cos^(-1)(z) an I cancelled out the i's but I've never ever been good with logs so not sure what I can work with

I also tried using sinx = [e^(iz)-e^(-ix)] / 2i divided by cosx = [e^(ix) + e^(-ix)]/2

I cancelled out a few things and ended up with xie^(iy) - 2e^(iy) + xie^(-y)

and I can't even do anything with that because there are no square powers so I can't factorise... so I know thats wrong too.. I've looked up how to derive the sin^(-1) and it was done in a similar fashion using sinx = [e^(iz)-e^(-ix)] / 2i and then letting k = e^(ix) and you end up with a quadratic and can use the Quadratic formula to get the answer... but I cant' see how that will work for tan^(-1) any info would be appreciated