Vtyjkyuk

If $x=cos a + i sin a$, $y=cos b + i sin b$, $z=cos c +i sin c$, and $x+y+z = xyz$, then show that $$cos(a-b) + cos(b-c) + cos(c-a) + 1=0$$

Here's how I tried it

$$x+y+z=xyz $$

So, by De Moivre's Theorem,
$$(cos a + cos b + cos c) + i(sin a + sin b + sin c) = cos(a+b+c) + i sin(a+b+c)
$$

Equating real and imaginary parts,
$$cos a + cos b + cos c= cos(a+b+c)$$
and similarly for sine. Now,

$$(a-b) + (b-c) + (c-a) =0$$

What to do now? Please help. And please use De Moivre Theorem!

Note that $$cos(a-b) = cos(a)cos(b)+sin(a)sin(b) = Re(xbary)$$
So $$labeleq1cos(a-b)+cos(b-c)+cos(c-a) = Re(xbary+ybarz+zbarx)$$
With that in mind, note that just like $x$, $y$ and $z$, the $xyz$ complex number is of modulus 1. So if we go back to the $x+y+z=xyz$ identity, we see that $$|x+y+z|=1$$
In other words, $$beginsplit
1 & = |x+y+z|^2 \
& = (x+y+z)(bar x+bar y +bar z) \
& = xbar x + x bar y + x bar z + ybar x + y bar y + y bar z + z bar x + z bar y + z bar z \
& = 1 + x bar y + x bar z + 1 + ybar x + y bar z + 1 + z bar x + z bar y \
& = 3 + (x bar y + y bar z + z bar x) + (ybar x + z bar y + x bar z) \
& = 3 + 2 Re(x bar y + y bar z + z bar x)
endsplit$$
We conclude that $$Re(xbary+ybarz+zbarx)+1=0$$ which yields the desired result.

(This is a followup on Stefan Lafon's answer, too long for a comment.)

Using that $,|x|=1 iff bar x = dfrac1x,$ and $,cos(a-b)= operatornameReleft(dfracxyright)=operatornameRe(x bar y) = operatornameRe(bar x y),$, it follows that:

$$
beginalign
cos(a-b) + cos(b-c) + cos(c-a) &= cos(a-b) + cos(a-c) + cos(b-c) \
&= operatornameReleft(xbar y+ x bar z + y bar zright) \
&= operatornameReleft(xleft(bar y + bar zright) + y bar zright) \
&= operatornameReleft(xleft(bar x bar y bar z - bar xright) + y bar zright) \
&= operatornameReleft(bar y bar z + ybar z -1 right) \
&= operatornameReleft(left(y + bar yright) bar z -1 right) \
&= -1 + 2 operatornameRe(y) operatornameRe(z)
endalign
$$

For the latter RHS to be $,-1,$, the second term must be zero, so one of $,y,z,$ must be a purely imaginary number of modulus $,1,$ i.e. $,pm i,$. With some more legwork, it follows that the solution set of the given constraints is $, = 1,$ or permutations thereof.

Squaring & adding we get

$$cos^2(a+b+c)+sin^2(a+b+c)=(cos a+cos b+cos c)^2+(sin a+sin b+sin c)^2$$

$$implies1=1+1+1+2sum_textcyc(a,b,c)cos(a-b)$$

Observe that we actually don't need $x+y+z=xyz$

The sufficient condition is $$ x+y+z=cos p+isin piff|x+y+z|=1$$