Question

If $f : C \to C$ is defined by $f(x) = x^3$ and $\omega$ is a cube root of unity , then $f^{-1} (64) = $

• A. $\{4, 4 \omega , 4 \omega^2\}$
• B. $\{ 4 \}$
• C. $\{ 4, 4\omega \}$
• D. $\phi$

Answer 1

Answer: (A)

We have, $f (x) = x^3$
Let $f^{-1}(x) = y$
$\Rightarrow \, x = f (y) = y^3 \Rightarrow \, y = x^{1/3} $ or $f^{-1} (x) = x^{1/3}$
$\therefore f^{-1} (64) = (64)^{1/3} = 4·(1)^{1/3}$
We know that $ (1)^{1/3}$ has $\{1, \omega, \omega^2\}$ roots.
$\Rightarrow \, f^{-1}(64) = 4·(1)^{1/3}$ has roots $\{4, 4\omega, 4 \omega^2\}$