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Question

Write $(\frac{1}{2}-\frac{i\sqrt{3}}{2})^{12}$ in $a+bi$ form

My thoughts:

My initial thoughts was to convert into $R \;\text{cis}(\theta)$ form in which i got $\theta$ to be $-\frac{\pi}{3}$ but I do not know how to proceed from here

I attempted to go further by doing the following:

$(cos\frac{-\pi}{3}+i\sin\frac{-\pi}{3})^{12}$

to get $\frac{1}{2^{12}}-i\frac{\sqrt3}{2}$

  • 3
    After converting into $R \operatorname{cis} \theta$ form, use the De Moivre formula – 2017-02-03
  • 1
    $\frac {i\sqrt 2}{2}$ or $\frac {i\sqrt 3}{2}$ – 2017-02-03

2 Answers 2

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I think there is a typo in the question. As we have $\cos (-\frac {\pi }{3}) = -0.5$, then the term associated with the imaginary part should be $\sin -\frac {\pi}{3} = -\frac {\sqrt {3}}{2} $if you want to use De Moivre's formula.

Considering this done, use the formula which states that $$(\cos x + i\sin x )^n = \cos nx + \sin nx$$ We have $$(\cos (-\frac {\pi}{3}) + i\sin (-\frac {\pi}{3}))^{12} $$ Can you proceed from here?

If your question is correct, you can proceed with Arnaldo's method. Hope it helps.

  • 0
    typo is now fixed – 2017-02-03
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Hint

$$(1-i\sqrt{2})^2=-1-2\sqrt{2}i\to (1-\sqrt{2}i)^4=(1+2\sqrt{2}i)^2=-7+4\sqrt{2}i$$

Now calculate:

$$(-7+4\sqrt{2}i)^3$$