Let$z$ be a complex number such that $$ z + \frac{1}{z} = \cos(x) $$ Then what is the value of the expression $$ z^n + \frac{1}{z^n} $$ where $n$ is an integer?

Please help me, i have tried somehow using the trigonometric way of defining complex numbers but still didn't manage to get anywhere.

  • $\begingroup$ What do you mean by $z+\frac{1}{z}=\cos(x)?$ $\endgroup$ – Chickenmancer Mar 14 at 18:58
  • $\begingroup$ Is $z=x+iy$ and the $x$ in $\cos(x)$ refers to that?? $\endgroup$ – Anurag A Mar 14 at 19:01
  • $\begingroup$ z+1/z = cos(x) it's a relation that needs to help us to get to the other answer. $\endgroup$ – Melinceanu Victor Mar 14 at 19:04
  • $\begingroup$ so what is $x$? How is it related to $z$? $\endgroup$ – Anurag A Mar 14 at 19:07
  • 2
    $\begingroup$ You know, this would be a whole lot nicer if that were $z+\frac1z=2\cos x$... $\endgroup$ – jmerry Mar 14 at 19:46


Let $$z = \cos x + i \sin x$$ and $$\dfrac {1}{z} = \cos x - i \sin x$$

Then by DeMoivre's theorem $$z^n = (\cos x + i \sin x)^n = \cos nx + i \sin nx$$ $$\dfrac {1}{z^n} = (\cos x - i \sin x)^n = \cos nx - i \sin nx$$

What do you need to get $z^n + \dfrac {1}{z^n}?$

  • $\begingroup$ Here $z+\frac 1z=2\cos x$ but OP said $z+\frac 1 z=\cos x$ $\endgroup$ – J. W. Tanner Mar 14 at 23:01
  • $\begingroup$ I just read jmerry's comment above...I would think the sentiment would be the same. $\endgroup$ – bjcolby15 Mar 14 at 23:22

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