How do you convert $(- 2 \sqrt{3}) - 2 i$ $(-2sqrt3) - 2i$ to polar form?

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How do you convert $(- 2 \sqrt{3}) - 2 i$ $(-2sqrt3) - 2i$ to polar form?

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Answer 1 · 2017-04-18T08:25:37

$- 2 \sqrt{3} - 2 i = 4 (cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3})) = 4 e^{i \frac{4 π}{3}}$ $-2sqrt3-2i=4(cos((4pi)/3)+isin((4pi)/3))=4e^(i(4pi)/3)$

Explanation:

Modulus or absolute value of $- 2 \sqrt{3} - 2 i$ $-2sqrt3-2i$ is

$\sqrt{{(- 2 \sqrt{3})}^{2} + {(- 2)}^{2}}$ $sqrt((-2sqrt3)^2+(-2)^2)$

= $\sqrt{4 \times 3 + 4} = \sqrt{16} = 4$ $sqrt(4×3+4)=sqrt16=4$

Hence $- 2 \sqrt{3} - 2 i$ $-2sqrt3-2i$

= $4 (- \frac{2 \sqrt{3}}{4} - \frac{2}{4} i)$ $4(-(2sqrt3)/4-2/4i)$

= $4 (- \frac{\sqrt{3}}{2} - \frac{1}{2} i)$ $4(-sqrt3/2-1/2i)$

= $4 (cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3})) = 4 e^{i \frac{4 π}{3}}$ $4(cos((4pi)/3)+isin((4pi)/3))=4e^(i(4pi)/3)$

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How do you convert $(- 2 \sqrt{3}) - 2 i$ $(-2sqrt3) - 2i$ to polar form?

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