trigonometry?

Question

how do you find (1 - i√3)(√3 + 1) ?

and how do you convert (2, -2√3) in polar coords?
[ r ≥ 0 and θ between 0˚ and 360˚ ]

Answer 1

Hi,

The first one is ugly. I wonder if it's a typo. Either way, to eliminate the radical on the bottom, you must multiply by the conjugate of the bottom, radical 3 - 1, on both the top and bottom of the fraction. When you do, you have:

(1-i*radical3)(radical3 -1)
-------------------------------------
(radical3+1)(radical3-1)

This multiplies out to

radical3-1-3i+i*radical3
----------------------------------
3 -1

which simplifies slightly to

radical3-1-3i+i*radical3
----------------------------------
2

Second problem

If you graph (2,-2radical3) you get a point in the 4th quadrant that has a reference angle of 60 degrees, so the angle is 360 - 60 or 300 degrees. Using the distance formula or the pattern for sides for a 30-60-90 triangle, the hypotenuse has a length of 4. So polar form for this is (4,300) using degrees.

I hope this helps.

Answer 2

If question is (1 - √3i).(√3 + 1)
= √3 + 1 - 3i -√3.i
= (√3 + 1) - (3 + √3).i
However if question was
(1 - √3.i).(√3i + 1)
= √3i + 1 - 3i² - √3i
= 1 + 3
= 4 (just a thought!)

Question 2
coordinates are ( 2,-2/√3) ie in 4th quadrant.
Angle in 4th quadrant = a say
tan a = 2/√3 / 2 = 1/√3
a = 30° from horizontal in 4th quadrant.
polar coordinates are 2 at angle 330° or
2 at angle 11π/6