Verifying and Proving z=cisθ Identity

[danago]

If z=cis\theta, verify that <br /> \tan \theta = \frac{{z - z^{ - 1} }}{{i(z + z^{ - 1} )}}<br />. Use this result to prove that <br /> \cos (2\theta ) = \frac{{1 - \tan ^2 \theta }}{{1 + \tan ^2 \theta }}<br />


Ok, I've managed to verify the first equation given, but I am not really sure how to use it to prove the second identity. I am really not sure where to start. If somebody could give me a hint about where to start id be very appreciative.

Thanks,
Dan.

 

[mjsd]

Does your z^{-1} actually mean conjugate of z?

 

[danago]

Nah its the reciprocal of z.

 

[HallsofIvy]

Haven't seen "cis" in a long time! It's engineering shorthand for cos(\theta)+ i sin(\theta)= e^{i\theta}. From the last form, or comparing \theta for z and z^-1, it should be clear that if z= cis(\theta) then z^{-1}= cis(-\theta)= cos(\theta)- i sin(\theta). Putting those in for z and z^-1 in
\frac{z- z^{-1}}{i(z+ z^{-1}}[/itex]<br /> <br /> By the way, when |z|= 1, as is the case here, z^-1 <b>is</b> the complex conjugate:<br /> \frac{1}{x+ iy}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+ y^2}= x- iy

 

[mjsd]

basically, you have \tan \theta equal something, so all you need to do to prove your 2nd identity is just sub in this where \tan \theta's appear and simplify then you shall see that it actually equals to \cos (2\theta)
hint:
\cos (x)=\frac{e^{ix}+e^{-ix}}{2}

 

[mjsd]

Nah its the reciprocal of z.

Nah, my question was a rhetorical one

also note that z+z^*=2\text{Re}(z) , z-z^* = 2i\text{Im}(z) and z z^* = |z|

 

[danago]

Got it thanks for the help guys :D