Imaginary numbers and real numbers?

[Ilikebugs]

z is either a real, imaginary or complex number, and z^12=1 and z^20 also equals 1. What are all possible values of z?

I know 1 and -1 are them, and I think its also i and -i?

 

[MarkFL]

We have:

\(\displaystyle z^{12}=1=z^{20}\)

We know $z\ne0$ so we may divide through by $z^{12}$ to get:

\(\displaystyle z^8-1=0\)

\(\displaystyle \left(z^4-1\right)\left(z^4+1\right)=0\)

Now, we observe that $z^4=-1$ satisfies neither original equation, so we are left with:

\(\displaystyle \left(z^4-1\right)=0\)

Next, factor as the difference of squares:

\(\displaystyle \left(z^2+1\right)\left(z^2-1\right)=0\)

Since $1=-(-1)=-i^2$, we may write:

\(\displaystyle \left(z^2-i^2\right)\left(z^2-1\right)=0\)

Factor again:

\(\displaystyle (z+i)(z-i)(z+1)(z-1)=0\)

Thus, $z\in\{-i,+i,-1,1\}$, as you thought. :D