Question

If $z$ is a complex number satisfying $z+z^{-1}=1$, then $z^n+z^{-n},nϵN$ has the value

• A. $2(-1{)}^n$ when $n$ is a multiple of $3$
• B. $(-1{)}^{n-1}$ when $n$ is not a multiple of $3$
• C. $(-1{)}^{n+1}$ when $n$ is a multiple of $3$
• D. $0$ when $n$ is not a multiple of $3$

Answer 1

Answer: (A), (B)

$2(-1{)}^n$ when $n$ is a multiple of $3$
$(-1{)}^{n-1}$ when $n$ is not a multiple of $3$

$z=|z|(\cos \theta +i\sin \theta )$

Hence, $z^{-1}=\overline{z}=|z|(\cos \theta -i\sin \theta )$

Therefore, $z+\overline{z}$$=2|z|\cos \theta$$=1$

$\cos \theta =\frac{1}{2|z|}$

Considering $|z|=1$, we get

$\cos \theta =\frac{1}{2}$$\theta =\frac{\pi }{3}$.

Now $z^n+\overline{z^n}$

$=2\cos n\theta$

$=2\cos \left(\frac{n\pi }{3}\right)$

If $n$ is multiple of $3$,

Case I: $n$ is an even multiple, we get

$2\cos \left(2k\pi \right)$

$=2$

Case II: $n$ is an odd multiple, we get

$2\cos \left(\right(2k-1\left)\pi \right)$

$=-2$

Hence, if $n$ is multiple of $3$, we get

$2(-1{)}^n$

Else if $n$ is not a multiple of $3$, we get

$(-1{)}^{n-1}$