We are looking for an alternative proof for the following system of equations:
$t$ and $s$ are two complex numbers and $\phi$ a real number such that $$\Sigma:\,\begin{cases} t^2+s^2=1\\ (s+it)^n+(s-it)^n=2\cos(\phi) \end{cases} $$ Here’s one approach:
Since $s^2+t^2=1$, $ s+it$ and $s-it$ are conjugate therefore there is $\alpha$ so that $s+it=e^{i\alpha}$ and $s-it=e^{-i\alpha}$
Edit:( $\color\red{\text{this last point is false as pointed by Marty Cohen}})$.
(Btw it should be a way of proving that: if $\Sigma$ holds then $s$ and $t$ are real numbers.)
Thus $\Sigma$ is equivalent to: $\cos{(n\alpha)}=\cos\phi$ which, modulo $2\pi$ provides two solutions: $\alpha=\dfrac{\phi}{n}$ and $\alpha=-\dfrac{\phi}{n}$.
Finally: $$\Sigma\iff \begin{cases} s+it=\exp\left(\dfrac{\phi}n\right)\\ s-it =\exp\left(-\dfrac{\phi}n\right) \end{cases} \textbf{ or } \begin{cases} s+it=\exp\left(-\dfrac{\phi}n\right)\\ s-it =\exp\left(\dfrac{\phi}n\right) \end{cases} $$ Which is easy to solve.
You say that $s$ and $t$ are complex numbers, not just real numbers.
If $s = a+ib, t = c+id$, then $s+it =a+ib+i(c+id) =a-d+i(b+c) $ and $s-it =a+ib-i(c+id) =a+d+i(b-c) $ so they are not conjugate.
However $s^2+t^2 =(a+ib)^2+(c+id)^2 =a^2-b^2+c^2-d^2+2i(ab+cd) $ and if this $=1$ then $a^2-b^2+c^2-d^2 =1 $ and $ab+cd = 0$.
Going a little further
$\begin{array}\\ r_n &=(s+it)^n+(s-it)^n\\ &=(a-d+i(b+c))^n+(a+d+i(b-c))^n\\ &=\sum_{k=0}^n \binom{n}{k}i^{n-k}((a-d)^k(b+c)^{n-k}+(a+d)^k(b-c)^{n-k})\\ \text{so}\\ r_{2n-1} &=(s+it)^{2n-1}+(s-it)^{2n-1}\\ &=\sum_{k=0}^{2n-1} \binom{2n-1}{k}i^{2n-1-k}((a-d)^k(b+c)^{2n-1-k}+(a+d)^k(b-c)^{2n-1-k})\\ &=\sum_{k=0}^{n-1} \left(\binom{2n-1}{2k}i^{2n-1-2k}((a-d)^{2k}(b+c)^{2n-1-2k}+(a+d)^{2k}(b-c)^{2n-1-2k})+\binom{2n-1}{2k+1}i^{2n-1-2k-1}((a-d)^{2k+1}(b+c)^{2n-1-2k-1}+(a+d)^{2k+1}(b-c)^{2n-1-2k-1})\right)\\ &=\sum_{k=0}^{n-1} \left(\binom{2n-1}{2k}(-i)(-1)^{n-k}((a-d)^{2k}(b+c)^{2n-1-2k}+(a+d)^{2k}(b-c)^{2n-1-2k})+\binom{2n-1}{2k+1}(-1)^{n-k-1}((a-d)^{2k+1}(b+c)^{2n-1-2k-1}+(a+d)^{2k+1}(b-c)^{2n-1-2k-1})\right)\\ &=(-i)\sum_{k=0}^{n-1} \binom{2n-1}{2k}(-1)^{n-k}((a-d)^{2k}(b+c)^{2n-1-2k}+(a+d)^{2k}(b-c)^{2n-1-2k})\\ &+\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}(-1)^{n-k-1}((a-d)^{2k+1}(b+c)^{2n-1-2k-1}+(a+d)^{2k+1}(b-c)^{2n-1-2k-1})\\ \end{array} $
And that's enough for now.