How do you use demoivre's theorem to simplify 2(3+i)2?

First convert your z=3+i into polar form by taking its absolute value and argument.Recall that |z|=(3)3+12=2.And that arg(z)=arctan⁡(13)=π6∴z=2⋅cis(π6).By de Moivre's theorem, zn=rn⋅cis(nθ).∴z2=4⋅cis(π3).Converting back into Cartesian form, we have x=r⋅cos⁡(θ)and y=r⋅sin⁡(θ)x=4⋅cos⁡(π3)and y=4⋅sin⁡(π3).z2=2+23i.2z2=4+43i.

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Selena Cowan

Answered question

2022-01-31

How do you use demoivre's theorem to simplify

2 {(\sqrt{3} + i)}^{2}

Answer & Explanation

oferenteoo

Beginner2022-02-01Added 12 answers

First convert your

z = \sqrt{3} + i

into polar form by taking its absolute value and argument.
Recall that

| z | = \sqrt{{(\sqrt{3})}^{3} + 1^{2}} = 2

.
And that

a r g (z) = \arctan (\frac{1}{\sqrt{3}}) = \frac{π}{6}

∴ z = 2 \cdot c i s (\frac{π}{6})

.
By de Moivre's theorem,

z^{n} = r^{n} \cdot c i s (n θ)

∴ z^{2} = 4 \cdot c i s (\frac{π}{3})

.
Converting back into Cartesian form, we have

x = r \cdot \cos (θ) and y = r \cdot \sin (θ)

x = 4 \cdot \cos (\frac{π}{3}) and y = 4 \cdot \sin (\frac{π}{3})

z^{2} = 2 + 2 \sqrt{3} i

2 z^{2} = 4 + 4 \sqrt{3} i

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