How do you convert $5 \sqrt{3} - 5 i$ $5 sqrt 3 - 5i$ to polar form?

Question

12118 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-30T06:44:35

The answer is $= 10 (cos (- \frac{π}{6}) + i sin (- \frac{π}{6}))$ $=10(cos(-pi/6)+isin(-pi/6))$

Let $z = 5 \sqrt{3} - 5 i$ $z=5sqrt3-5i$

We must transform this equation to the form

$z = r (cos θ + i sin θ)$ $z=r(cos theta+isintheta)$

We calculate the modulus of $z$ $z$

$∣ ∣ z ∣ = \sqrt{25 \cdot 3 + 25} = \sqrt{100} = 10$ $∣z∣=sqrt(25*3+25)=sqrt100=10$

$z = ∣ z ∣ (\frac{5 \sqrt{3}}{∣ z ∣} - i \frac{5}{∣ z ∣})$ $z=∣z∣((5sqrt3)/(∣z∣)-i5/(∣z∣))$

$z = 10 (5 \frac{\sqrt{3}}{10} - i \frac{5}{10})$ $z=10(5sqrt3/10-i5/10)$

$z = 10 (\frac{\sqrt{3}}{2} - \frac{1}{2} i)$ $z=10(sqrt3/2-1/2i)$

Therefore,

$cos θ = \frac{\sqrt{3}}{2}$ $costheta=sqrt3/2$ and $sin θ = - \frac{1}{2}$ $sintheta=-1/2$

We are in the 4th quadrant

$θ = - \frac{π}{6}$ $theta=-pi/6$

Therefore,

$z = 10 (cos (- \frac{π}{6}) + i sin (- \frac{π}{6}))$ $z=10(cos(-pi/6)+isin(-pi/6))$

$z = e^{\frac{- i π}{6}}$ $z=e^((-ipi)/6)$

How do you convert 5 sqrt 3 - 5i5√3−5i5 sqrt 3 - 5i to polar form?