How do you use demoivre's theorem to simplify $2(sqrt3+i)^2$ ?

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Answer 1 · 2017-09-11T08:38:29

$2(sqrt(3) + i)^2 = 4 + 4sqrt(3)i$

First convert your $z = sqrt(3) + i$ into polar form by taking its absolute value and argument.

Recall that $|z| = sqrt((sqrt(3))^2 + 1^2) = 2$ .

And that $Arg(z) = arctan(1/sqrt(3)) = pi/6$ .

$:. z = 2* cis(pi/6)$ .

By de Moivre's theorem, $z^n = r^n*cis(ntheta)$ .

$:. z^2 = 4*cis(pi/3)$ .

Converting back into Cartesian form, we have $x=r*cos(theta)$ and $y=r*sin(theta)$

$:. x=4*cos(pi/3)$ and $y=4*sin(pi/3)$ .

$:. z^2=2+2sqrt(3)i$ .

$:. 2z^2 = 4+4sqrt(3)i$ .

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