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

  1. Mar 21, 2007 #1
    I have to prove that

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


    2. Relevant equations

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


    3. 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
     
  2. jcsd
  3. Mar 21, 2007 #2

    Dick

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    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.
     
    Last edited: Mar 21, 2007
  4. Mar 21, 2007 #3

    Dick

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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?
     
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