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kuhse4461a

Answered question

2022-01-17

How do I solve

z^{2} = 2 + 2 r t 3 i

? For a quadratic I find

y^{2} - r t 2 y + 1 r t 3

but my answer is nothing like a solve. Can someone advise?

Ana Robertson

Beginner2022-01-18Added 26 answers

\sqrt{2^{2} + (2 \sqrt{3}) 2} = \sqrt{4 + 12} = \sqrt{16} = 4

a r c \tan (2 \frac{\sqrt{3}}{2}) = a r c \tan (\sqrt{3} = \frac{π}{2}

z^{2} = (4) (\frac{π}{3})

in polar form

z = 2 ((\frac{π}{6}) + k π) for k = 0 or k = 1

z = 2 (\frac{π}{6}) = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6})) = 2 (\frac{\sqrt{3}}{2} + i (\frac{1}{2}) = \sqrt{3} + i

z = 2 (\frac{7 π}{6}) = 2 (\cos (\frac{7 π}{6}) + i \sin (\frac{7 π}{6}) = 2 (\frac{\sqrt{3}}{2} - i ((\frac{1}{2}) = - \sqrt{3} - i

yotaniwc

Beginner2022-01-19Added 34 answers

z^{2} = 2 + 2 \sqrt{3} i = 4 [0.5 + 0.5 \sqrt{3} i] = 4 [\cos (\frac{π}{3}) + i \sin (\frac{π}{3})] = 4 e \frac{i π}{3}

Taking square roots:

z = \pm \sqrt{4} e \frac{i π}{6} = \pm 2 e \frac{i π}{6} = \pm 2 [\cos (\frac{π}{6}) + i \sin (\frac{π}{6})]

= \pm 2 [0.5 \sqrt{3} + 0.5 i] = \pm [\sqrt{3} + i]

Answers: 1.

z = \sqrt{3} + i

z = - \sqrt{3} - i

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