# How to derive the formula for tan(-1)(z) ~ Complex variables

• laura_a
In summary, to prove that tan^(-1)(z) = i/2 * log[(i+z)/(i-z)], you can use the fact that exp(i*z)=cos(z)+i*sin(z) for complex z and let q=exp(i*z). By writing out the formula for tan(z) in terms of q and rearranging, we can solve for q^2 and then use the definition of q to get the desired result.

#### laura_a

I have to prove that

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 canceled 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 canceled 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 that's 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

The trick to doing these is to remember that exp(i*z)=cos(z)+i*sin(z) even if z is complex. So you can write tan(z)=(-i)*(exp(i*z)-exp(-i*z))/(exp(i*z)+exp(-i*z)). Now when you realize exp(-i*z)=1/exp(i*z), inverting it is just solving a quadratic equation.

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Ok, take q=exp(i*z).

tan(z)=(-i)*(q-1/q)/(q+1/q)

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

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

Now q^2=exp(2*i*z). Can you take it from there?