How do you convert $\sqrt{3} - \sqrt{3} i$ $sqrt3 - sqrt3i$ to polar form?

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How do you convert $\sqrt{3} - \sqrt{3} i$ $sqrt3 - sqrt3i$ to polar form?

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Answer 1 · 2016-10-22T11:34:16

The polar form is $\sqrt{3} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $sqrt3(cos(-pi/4)+isin(-pi/4))$

Explanation:

Let $z = \sqrt{3} - i \sqrt{3}$ $z=sqrt3-isqrt3$

Then the modulus of z is $∣ ∣ ∣ z ∣ = \sqrt{{(\sqrt{3})}^{2} + {(\sqrt{3})}^{2}} = \sqrt{6}$ $∣z∣=sqrt((sqrt3)^2+ (sqrt3)^2) =sqrt6$
Rewriting z

$z = ∣ z ∣ (\sqrt{3} - i \sqrt{3}) = \sqrt{6} ((\frac{\sqrt{3}}{\sqrt{6}}) - i (\frac{\sqrt{3}}{\sqrt{6}}))$ $z=∣z∣(sqrt3-isqrt3)=sqrt6((sqrt3/sqrt6)-i(sqrt3/sqrt6))$

$= \sqrt{6} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})$ $=sqrt6(1/sqrt2-i/sqrt2)$

Comparing this to the standard form $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

We see that $cos θ = \frac{1}{\sqrt{2}}$ $costheta=1/sqrt2$

and $sin θ = - \frac{1}{\sqrt{2}}$ $sintheta=-1/sqrt2$
This can occur in the 4th quadrant and $θ = - \frac{π}{4}$ $theta =-pi/4$

And finally we have $z = \sqrt{6} (cos (- \frac{π}{4}) + sin (- \frac{π}{4}))$ $z=sqrt6(cos(-pi/4)+sin(-pi/4))$

You can also write the resultas $z = \sqrt{6} e^{- i \frac{π}{4}}$ $z=sqrt6e^(-ipi/4)$

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How do you convert $\sqrt{3} - \sqrt{3} i$ $sqrt3 - sqrt3i$ to polar form?

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Explanation:

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