How do you convert $\sqrt{3} i - 1$ $sqrt(3)i - 1$ to polar form?

Question

How do you convert $\sqrt{3} i - 1$ $sqrt(3)i - 1$ to polar form?

1 Answer

Impact of this question

2164 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T20:53:18

$(2, \frac{2 π}{3})$ $(2,(2pi)/3)$

Explanation:

Using the formulae that links Cartesian to Polar coordinates.

$∙ r = \sqrt{x^{2} + y^{2}} and θ = {tan}^{- 1} (\frac{y}{x})$ $•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)$

now $\sqrt{3} i - 1 = - 1 + \sqrt{3} i$ $sqrt3 i-1=-1+sqrt3 i$

here x = -1 and y = $\sqrt{3}$ $sqrt3$

$\Rightarrow r = \sqrt{{(- 1)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{4} = 2$ $rArrr=sqrt((-1)^2+(sqrt3)^2)=sqrt4=2$

and $θ = {tan}^{- 1} (- \sqrt{3}) = - \frac{π}{3}$ $theta=tan^-1(-sqrt3)=-pi/3$

$- 1 + \sqrt{3} i is in the 2nd quadrant$ $-1+sqrt3 i" is in the 2nd quadrant"$

so $θ requires to be an angle in 2nd quadrant$ $theta" requires to be an angle in 2nd quadrant"$

$\Rightarrow θ = (π - \frac{π}{3}) = \frac{2 π}{3}$ $rArrtheta=(pi-pi/3)=(2pi)/3$

$\Rightarrow (- 1, \sqrt{3}) = (2, \frac{2 π}{3})$ $rArr(-1,sqrt3)=(2,(2pi)/3)$

Science

Math

Humanities

... and beyond

How do you convert $\sqrt{3} i - 1$ $sqrt(3)i - 1$ to polar form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you convert √3i−1sqrt(3)i - 1 to polar form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you convert $\sqrt{3} i - 1$ $sqrt(3)i - 1$ to polar form?