Chapter 6.5, Problem 61HP

To determine

To Prove:

  $\frac{-1-i\sqrt{3}}{2}$ is the cube root of 1.

Explanation of Solution

Concept Used :

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• $\begin{array}{l}\sqrt{-1}=i\\ -1=i^2\end{array}$
• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

• Like radical expressions : If the radicals have both the index and radicand identical , then they are like radicals.
• In order to add or subtract , the radicals must be like radicals.

Calculation:

$\begin{array}{l}\text{To show : }\\ \frac{-1-i\sqrt{3}}{2}=\sqrt[3]{1}\\ \text{That is , to show}\\ {\left(\frac{-1-i\sqrt{3}}{2}\right)}^3=1\\ \text{Consider, }\\ {\left(\frac{-1-i\sqrt{3}}{2}\right)}^3\\ =\left(\frac{-1-i\sqrt{3}}{2}\right)\cdot \left(\frac{-1-i\sqrt{3}}{2}\right)\cdot \left(\frac{-1-i\sqrt{3}}{2}\right)\\ =\frac{\left(-1-i\sqrt{3}\right)\left(-1-i\sqrt{3}\right)\left(-1-i\sqrt{3}\right)}{8}\mathrm{...............}[\text{Product property of radicals]}\\ =\frac{\left(1+i\sqrt{3}+i\sqrt{3}+3i^2\right)\left(-1-i\sqrt{3}\right)}{8}\\ =\frac{\left(1+2i\sqrt{3}-3\right)\left(-1-i\sqrt{3}\right)}{8}\mathrm{..........................}[\text{Adding like radicals]}\\ \text{=}\frac{\left(2i\sqrt{3}-2\right)\left(-1-i\sqrt{3}\right)}{8}\\ =\frac{-2i\sqrt{3}-6i^2+2+2i\sqrt{3}}{8}\\ =\frac{(2i-2i)\sqrt{3}+8}{8}\mathrm{........................................}[\text{Adding like radicals]}\\ \text{=}\frac{8}{8}=1\end{array}$

So, ${\left(\frac{-1-i\sqrt{3}}{2}\right)}^3=1$

Hence $\frac{-1-i\sqrt{3}}{2}$ is the cube root of 1.