Understanding a Complex Math Equality: Cosine and Cubic Functions Explained

[Mentallic]

Homework Statement

I found this equality in the thread https://www.physicsforums.com/showthread.php?t=407130"
<br /> cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}} <br />
and I'd like to know how it works.

Note: I haven't studied this, but I do know about complex numbers and I got some hints from that thread on what to do.

The Attempt at a Solution

First of all, I assumed it was true.

For the left side of the equality, let y=cos\left(\frac{1}{3}arccos(z)\right)

so z=cos\left(3arccos(y)\right)

by trig identities, cos\left(3arccos(y)\right)=4y^3-3y

So the solutions to y=cos\left(\frac{1}{3}arccos(z)\right) are the solutions (not exactly sure which of the 3) to the cubic 4y^3-3y-z=0

and now taking the right side, y=\frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}

z+\sqrt{z^2-1}=cos(z)+isin(z)

simplifying this gives y=cos(z/3)

So hence for this equality to be true, 4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-z=0 for all z, but this isn't the case.

Please help me understand this more

 

[tiny-tim]

Hi Mentallic!

So hence for this equality to be true, 4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-z=0 for all z, but this isn't the case.

You mean 4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-cos(z)=0, which is true …

just expand cos(z/3 + 2z/3).

 

[Mentallic]

But z=cos\left(3arccos(y)\right)

and cos\left(3arccos(y)\right)=4y^3-3y where y=cos\left(\frac{1}{3}arccos(z)\right)

So shouldn't it then be 4y^3-3y=z ?

Yeah if I change it to cos(z) it works, but I still don't see where I've gone wrong.

 

[tiny-tim]

You're interchangeably using z as an angle and as a cosine.

In particular, your …

z+\sqrt{z^2-1}=cos(z)+isin(z)

isn't true.

(unless cosz = z)

 

[Mentallic]

Oh of course...