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Question: Let \(f:C\to C\) be a complex-valued function defined by \(f\left( x \right)={{x}^{3}}\). Determine ...

Let f:C→Cf:C\to C be a complex-valued function defined by f(x)=x3f\left( x \right)={{x}^{3}}. Determine the set f−1(−1){{f}^{-1}}\left( -1 \right)

Explanation

Solution

Hint: Use the definition of f−1(−1){{f}^{-1}}\left( -1 \right) as \left\\{ x:{{x}^{3}}=1,x\in C \right\\}.Use the formula a3+b3=(a+b)(a2−ab+b2){{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) and zero product property. Use the quadratic formula for solving the quadratic equation generated while solving the question.

Complete step-by-step answer:

Let x∈f−1(−1)x\in {{f}^{-1}}\left( -1 \right)
Since {{f}^{-1}}\left( -1 \right)=\left\\{ x:{{x}^{3}}=1,x\in C \right\\} we have x3=−1{{x}^{3}}=-1
Adding 1 on both sides, we get
x3+1=0{{x}^{3}}+1=0
We know that a3+b3=(a+b)(a2−ab+b2){{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)
Using the above formula, we get
(x+1)(x2−x+1)=0\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)=0
From zero product property, we know that if ab = 0 then, a= 0 or b = 0.
Hence we have
x+1=0x+1=0 or x2−x+1=0{{x}^{2}}-x+1=0
If x + 1 = 0, then x = -1
If x2−x+1=0{{x}^{2}}-x+1=0 then we know that roots of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 are x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Here a = 1, b = -1 , c= 1
Using the above formula, we get
x=1±1−42=1±−32=1±i32x=\dfrac{1\pm \sqrt{1-4}}{2}=\dfrac{1\pm \sqrt{-3}}{2}=\dfrac{1\pm i\sqrt{3}}{2}
Taking + sign we get x=1+i32x=\dfrac{1+i\sqrt{3}}{2}
Taking – sign we get x=1−i32x=\dfrac{1-i\sqrt{3}}{2}
Hence we have {{f}^{-1}}\left( -1 \right)=\left\\{ -1,\dfrac{1+i\sqrt{3}}{2},\dfrac{1-i\sqrt{3}}{2} \right\\}

Note: Alternate solution 1.
We know that if x3=1{{x}^{3}}=1 then x=1,ω,ω2x=1,\omega ,{{\omega }^{2}} where ω=−1+i32\omega =\dfrac{-1+i\sqrt{3}}{2}
Now we have x3=−1{{x}^{3}}=-1
Put t = -x
So, we have t3=1{{t}^{3}}=1
Hence t=1,ω,ω2t=1,\omega ,{{\omega }^{2}}
Reverting to the original variable, we get
−x=1,ω,ω2 ⇒x=−1,−ω.−ω2 ⇒x=−1,1−i32,1+i32 \begin{aligned} & -x=1,\omega ,{{\omega }^{2}} \\\ & \Rightarrow x=-1,-\omega .-{{\omega }^{2}} \\\ & \Rightarrow x=-1,\dfrac{1-i\sqrt{3}}{2},\dfrac{1+i\sqrt{3}}{2} \\\ \end{aligned}
which is the same as obtained above.
Alternate Solution 2.
We know that ei(2k+1)π=−1,k∈Z{{e}^{i\left( 2k+1 \right)\pi }}=-1,k\in \mathbb{Z}
Hence we have

& {{x}^{3}}={{e}^{i\left( 2k+1 \right)\pi }} \\\ & \Rightarrow x={{e}^{i\dfrac{\left( 2k+1 \right)\pi }{3}}},k=0,1,2 \\\ \end{aligned}$$ Hence we have $$x={{e}^{\dfrac{\pi }{3}i}},{{e}^{\dfrac{3\pi }{3}i}},{{e}^{\dfrac{5\pi }{3}i}}$$ Using ${{e}^{ix}}=\cos x+i\sin x$, we get $x=\dfrac{1+i\sqrt{3}}{2},-1,\dfrac{1-i\sqrt{3}}{2}$ which is the same as obtained above Alternate Solution 3: If $1,\alpha ,{{\alpha }^{2}},\ldots ,{{\alpha }^{n-1}}$ are the nth roots of unity then the solutions of the equation ${{x}^{n}}=a,a\in R$ , such that there exists at least one real root, are $p,p\alpha ,p{{\alpha }^{2}},\ldots ,p{{\alpha }^{n-1}}$ where p is one of the real roots of the equations,e.g. $p={{a}^{\dfrac{1}{n}}}$ Taking n = 3 , a = -1 and $\alpha =\omega $ we get the desired result.