Let $z$ be a complex such that $|z-1| =1$, and consider the complex numbers $v$ and $w$ such that: $w = z^2 -z$ and $3\arg(v) = 2\arg(w)$, where arg is the argument of a complex number. Show that $$ \left({zv\over w}\right)^2$$ is a real number
I first tried to show that the imaginary part of the number is zero, knowing that half of the subtraction of a complex number and its conjugate is equal to its imaginary part. Nonetheless, i could not find a relation between $z$ and $v$, and I ended up in a answer with these two letters.
The answer for two non real numbers is possible as $i^2=-1$ but it isnt the case for $i^3=-i$ so we cant prove it for three non real numbers also if for a quadratic if roots are imaginary then they exist as conjugates and their product is a real number. Hope this helps you.Let the three numbers be $1+0i,2+0i,-3+0i$ so we directly get the result . I dont think there exists any rigorous way to prove it.REAL numbers are subset of complex numbers hence I have given you this example.