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Maths-

General

Easy

Question

The principal mod-amplitude form of $2 minus square root of 3 plus straight i$ is

$3 \sqrt{2} cis \frac{5 π}{12}$
$2 \sqrt{2} cis \frac{5 π}{12}$
$(\sqrt{6} - \sqrt{2}) cis \frac{5 π}{12}$
$2 \sqrt{3} cis \frac{π}{12}$

The correct answer is: $left parenthesis square root of 6 minus square root of 2 right parenthesis cis invisible function application fraction numerator 5 straight pi over denominator 12 end fraction$

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