Question

If $z$ is a complex number such that $z+z^{-1}=1$, then what is the value of $z^{99}+z^{-99}$?

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

The correct option is D -2

$z+\frac{1}{z}=3\Rightarrow \frac{z^2+1}{z}=1\Rightarrow z^2+1=z\Rightarrow z^2-z+1=0z=\frac{1\pm \sqrt{-3}}{2}=\frac{1\pm \sqrt{3}i}{2}$

$z=A\cdot e^{ib}\to$ exp. form

$A=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=1b={\tan }^{-1}(\pm \sqrt{3})=\pm \frac{\pi }{3}z=e^{\pm \pi /3}$

Now, $z^{99}={\left(e^{\pm \pi /3}\right)}^{99}=e^{\pm 33\pi }=e^{\pm \pi }=-1$

Similarly, $\frac{1}{z^{99}}=-1$

Answer: $=-1-1=-2$