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.

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asked by tabby
  1. 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 30Β° = tan Ο€/6 = 1/√3)
    but my angle is in IV, so Ø = 11Ο€/6 or 330Β°

    so 5√3 - 5i = 10(cos 11Ο€/6 + i sin 11Ο€/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 = Ο€ - acute angle
    - for III , Ø = Ο€ + acute angle
    - for IV, Ø = 2Ο€ - acute angle

    a + bi = r( cosØ + isinØ)

    use your calculator to verfiy your answer.

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    posted by Reiny

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