Complex number and exponent If $z = -1+ i\sqrt{3}$
Is it possible that to prove by using induction  $z^{2n}+2^n\cdot z^n+2^{2n}=0$ if $n$ is not multiple of $3$.
I know other way of proving it.
 A: Hint: $z^{3n}-2^{3n}=0$ by using Demoivre's theorem.
A: Note that $z^3=2^3$.  So, if $z^{2n}+2^nz^n+2^{2n}=0$ then
$$\eqalign{z^{2(n+3)}+2^{n+3}z^{n+3}+2^{2(n+3)}
  &=z^6z^{2n}+8z^32^nz^n+2^62^{2n}\cr
  &=2^6(z^{2n}+2^nz^n+2^{2n})\cr
  &=0\ .\cr}$$
If you check that the result is true for $n=1$ and $n=2$, then it follows by induction from the above that it is also true for $n=4,5,7,8,10,11,\ldots$, that is, for all $n$ not a multiple of $3$.
A: Where  Harish Chandra Rajpoot used induction on $2\cos(n\pi)\cos\left(\dfrac{n\pi}3\right)+1=0,$
 I'm tempted to use a little algebra 
Write $\dfrac{n\pi}3=x\iff n=?$  
$$2\cos x\cos3x+1=0$$
Using Werner's formula, $$\cos2x+\cos4x+1=0\iff\cos2x(2\cos2x+1)=0$$
If $\cos2x=0\implies2x=(2m+1)\dfrac\pi2$ where $m$ is any integer
$\implies n=\dfrac{3x}\pi=\dfrac{3(2m+1)}4$
If $2\cos2x+1=0\implies2x=2m\pi\pm\dfrac{2\pi}3=\dfrac{2\pi}3(3m\pm1)$
$\implies n=\dfrac{3x}\pi=3m\pm1$
A: Notice, $$z=-1+i\sqrt{3}=2\left(-\frac{1}{2}+i\frac{\sqrt 3}{2}\right)$$
$$=2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)$$
Now, we have $$z^{2n}+2^nz^n+2^{2n}=\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^{2n}+2^n\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^n+2^{2n}$$ $$=2^{2n}\left(\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}\right)+2^{2n}\left(\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}\right)+2^{2n}$$ $$=2^{2n}\left[\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}+\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}+1\right]$$ 
$$=2^{2n}\left[\left(\cos\frac{4n\pi}{3}+\cos\frac{2n\pi}{3}\right)+i\left(\sin \frac{4n\pi}{3}+\sin \frac{2n\pi}{3}\right)+1\right]$$ 
$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+2i\sin(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$ 
$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$ 
$\forall\quad n\neq 3m$ where $m$ is an integer.
step 1: setting $n=1$ in the above expression, we get $$2^{2}\left[2\cos(\pi)\cos\left(\frac{\pi}{3}\right)+1\right] $$ $$=4\left[2(-1)\frac{1}{2}+1\right]=0$$ Hence, the given equality is true for $n=1$


*Let's assume that it holds for $n=k$ then we get 


$$2^{2k}\left[2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)+1\right]=0$$
$$\iff 2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)=-\frac{1}{2}$$
3. Substituting $n=k+1$ in the above equality, we get 
$$=2^{2(k+1)}\left[2\cos((k+1)\pi)\cos\left(\frac{(k+1)\pi}{3}\right)+1\right]$$
$$=4\cdot2^{2k}\left[\cos\left(k\pi+\pi\right)\cos \left(\frac{k\pi}{3}+\frac{\pi}{3}\right)+1\right]$$
$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\cos\frac{\pi}{3}-\sin\frac{k\pi}{3}\sin\frac{\pi}{3}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\frac{1}{2}-\sin\frac{k\pi}{3}\frac{\sqrt 3}{2}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\frac{1}{2}\cos(k\pi)\cos\frac{k\pi}{3}+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$ substituting value from (2)
$$=4\cdot2^{2k}\left[-\frac{1}{2}\left(-\frac{1}{2}\right)+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$  $$=4\cdot2^{2k}\left[\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+\frac{5}{4}\right]\neq 0$$ Hence, we find from (1), (2) & (3) that the given equality doesn't hold for all positive integers $n\geq 2$
