If f: C → C is defined by f(x) = x3 and ω is a cube root of unity , then f-1 (64) =

Question

If f: C → C is defined by f(x) = x3 and ω is a cube root of unity , then f-1 (64) =  (A)  4, 4 ω , 4 ω2  (B)  4  (C)  4, 4ω  (D) φ

Tardigrade Admin · Accepted Answer

We have, f (x) = x3 Let f-1(x) = y ⇒ x = f (y) = y3 ⇒ y = x1/3 or f-1 (x) = x1/3 ∴ f-1 (64) = (64)1/3 = 4·(1)1/3 We know that (1)1/3 has 1, ω, ω2 roots. ⇒ f-1(64) = 4·(1)1/3 has roots 4, 4ω, 4 ω2

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

${4, 4 ω, 4 ω^{2}}$

${4}$

${4, 4 ω}$

$ϕ$

Q. If f:C→C is defined by f(x)=x3 and ω is a cube root of unity , then f−1(64)=

{4,4ω,4ω2}

{4}

{4,4ω}

ϕ

We have, f(x)=x3 Let f−1(x)=y ⇒x=f(y)=y3⇒y=x1/3 or f−1(x)=x1/3 ∴f−1(64)=(64)1/3=4⋅(1)1/3 We know that (1)1/3 has {1,ω,ω2} roots. ⇒f−1(64)=4⋅(1)1/3 has roots {4,4ω,4ω2}

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

${4, 4 ω, 4 ω^{2}}$

${4}$

${4, 4 ω}$

$ϕ$