Verifying and Proving z=cisθ Identity

[danago]

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


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
[tex]\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 /> $$\frac{1}{x+ iy}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+ y^2}= x- iy$$[/tex]

 

[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