Question

# How do you use DeMoivre's theorem to simplify (2+2i)^6?

How do you use DeMoivre's theorem to simplify (2+2i)^6?

Answer 1

$- 512 i$

#### Explanation:

Before applying $\textcolor{b l u e}{\text{De Moivre's theorem}}$ we requireto convert the complex number into trigonometric form.

To convert from $\textcolor{b l u e}{\text{complex to trigonometric form}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{z = x + y i = r \left(\cos \theta + i \sin \theta\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ where}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ and } \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

For 2 + 2i , we have x = 2 and y = 2

$\Rightarrow r = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{8} = 2 \sqrt{2}$

Now 2 + 2i is in the 1st quadrant and so we must ensure that $\theta$ is in the 1st quadrant.

$\Rightarrow \theta = {\tan}^{-} 1 \left(\frac{2}{2}\right) - {\tan}^{-} 1 \left(1\right) = \frac{\pi}{4} \text{ in 1st quadrant}$

$\Rightarrow 2 + 2 i = 2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) \textcolor{b l u e}{\text{ in trig form}}$

$\textcolor{b l u e}{\text{DeMoivre's theorem}}$ states that.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(r \left(\cos \theta + i \sin \theta\right)\right)}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)} |}}} n \in \mathbb{Q}$

$\Rightarrow {\left(2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)\right)}^{6} =$

${\left(2 \sqrt{2}\right)}^{6} \left(\cos \left(6 \times \frac{\pi}{4}\right) + i \sin \left(6 \times \frac{\pi}{4}\right)\right)$

$= 512 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(cos((3pi)/2)=0" and " sin((3pi)/2)=-1)color(white)(a/a)|)))

$\Rightarrow 512 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right) = 512 \left(0 - i\right)$

$\Rightarrow {\left(2 + 2 i\right)}^{6} = \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{- 512 i} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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