# Trig/PreCalc

Teach me how to express 5sqrt3 - 5i in polar form please.

I don't want you to do the work for me. Just show me the steps I need to do the work properly on my own. Otherwise I will not pass this class or the exam when I enter college, and I do not want to retake PreCalc.

Its a multiple choice homework question with answer choices:
a: 10(cos 11pi/6 + i sin 11pi/6)
b: 10(cos 11pi/6 - i sin 11pi/6)
c: 5 (cos 11pi/6 + i sin 11pi/6)
d: 10(cos 5pi/3 + i sin 5pi/3)

I did as much work as I could guess at, and came up with r=10sqrt2,
arctan=1/sqrt3 which equals sqrt3/3.I know that sqrt3/3 = tan 30 and 210 degrees. But now I am stuck.
Where do I go from here - or is this even the right way to try to solve this equation? Please help me.

1. π 0
2. π 0
3. π 44
1. start by graphing 5β3 - 5i
as (5β3, -5) in the argand plane
which would be in quadrant IV

r^2 = (5β3)^2 + (-5)^2 = 100
r = 10

we also know that tanΓ = -5/(5β3) = -1/β3
( I know that tan 30Β° = tan Ο/6 = 1/β3)
but my angle is in IV, so Γ = 11Ο/6 or 330Β°

so 5β3 - 5i = 10(cos 11Ο/6 + i sin 11Ο/6)

check:
RS = 10( β3/2 + i(-1/2)
= 5β3 - 5i
= original complex number

so in summary
for a + bi

1. sketch (a,b) to see which quadrant you are in
2. evaluate r,
with r^2 = a^2 + b^2 ---> r = | β(a^2 + b^2)
3. from tan Γ = |b/a| find the acute angel Γ
- for I, Γ is that acute angle
- for II , Γ = Ο - acute angle
- for III , Γ = Ο + acute angle
- for IV, Γ = 2Ο - acute angle

a + bi = r( cosΓ + isinΓ)

1. π 0
2. π 0
posted by Reiny

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