# Trig/PreCalc

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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.

• Trig/PreCalc -

start by graphing 5√3 - 5i
as (5√3, -5) in the argand plane
which would be in quadrant IV

r^2 = (5√3)^2 + (-5)^2 = 100
r = 10

we also know that tanØ = -5/(5√3) = -1/√3
( I know that tan 30° = tan π/6 = 1/√3)
but my angle is in IV, so Ø = 11π/6 or 330°

so 5√3 - 5i = 10(cos 11π/6 + i sin 11π/6)

check:
RS = 10( √3/2 + i(-1/2)
= 5√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 = | √(a^2 + b^2)
3. from tan Ø = |b/a| find the acute angel Ø
- for I, Ø is that acute angle
- for II , Ø = π - acute angle
- for III , Ø = π + acute angle
- for IV, Ø = 2π - acute angle

a + bi = r( cosØ + isinØ)

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