Let \(z\) be a complex number such that

\(z+\frac{1}{z}=1\)

What is \(z^3\)?

JovenlyCosmo Nov 20, 2023

#1**0 **

Squaring the equation z + 1/z = 1 gives z^2 + 2 + 1/z^2 = 1. Then multiplying by z^2 gives z^4 + 2z^2 + 1 = z^2. Rearranging, we get z^4 + 2z^2 + 1 - z^2 = 0, which factors as (z^2 + 1)(z^2 + 1) = 0. Since z^2 + 1 = 0 has no solutions, z^2 = -1. Taking the square root of both sides, we find z = i or z = -i.

If z = i, then z^3 = i^3 = -i.

Alternatively, if z = -i, then z^3 is also equal to -i.

Therefore, in either case, z^3 = -i.

bader Nov 20, 2023

#3**+1 **

Yeah, I managed to solve it 30 minutes later. (Sorry, the correct answer was -1.)

Anyways, thanks for the answer. What I did was take

\(z + \frac{1}{z}=1\) and multiply it by \(z^2\), getting \(z^3-z^2+z=0\). However, when you multiply \(z+ \frac{1}{z}=1\) by \(z\), you will get

\(z^2-z=1\)

From which you can replace the \(-z^2+z\) with \(-1\) (The negative of the equation above) to get

\(z^3-1=0\)

From which you can easily get \(z^3=-1\)

JovenlyCosmo Nov 21, 2023

#4**+1 **

Sorry I meant \(z^3+1=0\), and \(z^2-z=-1\).

Sorry if I confused anyone.

JovenlyCosmo
Nov 22, 2023