to simplify the complex number expression (−1/2 + sqrt(3)2i)³.

Question

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The problem asks to simplify the complex number expression $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3$ .
This expression is a complex cube root of unity, often denoted as $\omega$ . We know that $\omega^3 = 1$ .
Let's verify this using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ .

Let $a = -\frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}i$ .

Apply the binomial expansion formula.
$\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3 = \left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2\left(\frac{\sqrt{3}}{2}i\right) + 3\left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}i\right)^2 + \left(\frac{\sqrt{3}}{2}i\right)^3$
Calculate each term.
- First term:
  $\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$
- Second term:
  $3\left(-\frac{1}{2}\right)^2\left(\frac{\sqrt{3}}{2}i\right) = 3\left(\frac{1}{4}\right)\left(\frac{\sqrt{3}}{2}i\right) = \frac{3\sqrt{3}}{8}i$
- Third term: First, calculate $\left(\frac{\sqrt{3}}{2}i\right)^2$ :
  $\left(\frac{\sqrt{3}}{2}i\right)^2 = \frac{(\sqrt{3})^2}{2^2}i^2 = \frac{3}{4}(-1) = -\frac{3}{4}$
  Now, multiply by $3\left(-\frac{1}{2}\right)$ :
  $3\left(-\frac{1}{2}\right)\left(-\frac{3}{4}\right) = -\frac{3}{2} \cdot -\frac{3}{4} = \frac{9}{8}$
- Fourth term:
  $\left(\frac{\sqrt{3}}{2}i\right)^3 = \left(\frac{\sqrt{3}}{2}\right)^3 i^3 = \frac{3\sqrt{3}}{8}(-i) = -\frac{3\sqrt{3}}{8}i$
Combine all the calculated terms.
$-\frac{1}{8} + \frac{3\sqrt{3}}{8}i + \frac{9}{8} - \frac{3\sqrt{3}}{8}i$
Group the real and imaginary parts.
$\left(-\frac{1}{8} + \frac{9}{8}\right) + \left(\frac{3\sqrt{3}}{8}i - \frac{3\sqrt{3}}{8}i\right)$
Simplify the real and imaginary parts.
$\frac{-1+9}{8} + 0i$ $\frac{8}{8} + 0$ $1$

To simplify the complex number expression (-(1)/(2) + sqrt(3)2i)³.

Answer

Step-by-step solution

Apply the binomial expansion formula.

Calculate each term.

Combine all the calculated terms.

Group the real and imaginary parts.

Simplify the real and imaginary parts.

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