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I Proving De Moivre's Theorem for Negative Numbers?

  1. Apr 24, 2016 #1
    Or basically anything that isn't a positive integer.

    So I can prove quite easily by induction that for any integer n>0, De Moivre's Theorem (below) holds.

    If ##\DeclareMathOperator\cis{cis} z = r\cis\theta, z^n= r^n\cis(n\theta)##

    My proof below:

    Let P(n) be the statement ##z^n = r^n\cis(n\theta)##

    Prove true for n = 1:
    ##z^1 = z = r\cis\theta = r^1\cis(1\theta)##

    Assume true for n = k:
    ##z^k = r^k\cis(k\theta)##

    Prove true for n = k + 1:
    ##z^{k+1} = z^k \times z\\
    = r^k\cis(k\theta) * r\cis\theta\\
    = r^{k+1}\cis(k\theta)\cis\theta\\
    = r^{k+1}(\cos(k\theta) + i\sin(k\theta))((\cos\theta + i\sin\theta)\\
    = r^{k+1}(\cos(k\theta)\cos\theta + \cos(k\theta)i\sin\theta + i\sin(k\theta)\cos\theta - \sin(k\theta)\sin\theta)##
    then by compound angle identities
    ##= r^{k+1}(\cos(k\theta + \theta) + i\sin(k\theta + \theta))\\
    = r^{k+1}\cis((k+1)\theta)##

    However I struggle to do this with negatives. I tried the simple case of

    ##\frac{1}{r\cis\theta}## and from there, try to get to ##\frac{1}{r}\cis(-\theta)##, but I've only managed to do that by replacing the numerator of 1 with ##r\cis0##, which is equivalent, and then from there using De Moivre's Theorem to get an argument of ##-\theta##, which is... not very useful considering De Moivre's Theorem is what I'm trying to prove in the first place.

    I just can't manipulate it into a form that would allow me to use compound angle identities like I could with positive numbers. Any ideas?
     
  2. jcsd
  3. Apr 24, 2016 #2

    Math_QED

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    You can use $$z^{-n} = (z^n)^{-1}$$
     
  4. Apr 24, 2016 #3

    Samy_A

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    What is the relation between ##\frac{1}{\cis \theta}## and ##\cis (-\theta)##?
     
  5. Apr 24, 2016 #4
    Yeah, but I can't prove that that last step (raising the complex number to a power of -1) will make the angle/argument inside ##\cis## negative

    I know that ##\frac{1}{r\cis\theta} = (r\cis\theta)^{-1}##, but -1 is outside the boundaries for n that I have proved in my first post :(

    I know it works, I would just like to prove it.
     
  6. Apr 24, 2016 #5

    Math_QED

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    What is $$1/z$$ equal to?
     
  7. Apr 24, 2016 #6

    Samy_A

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    So what?
    Where in your proof for positive ##n## did it matter whether the angle is positive or negative?
    You didn't answer the question: what is the relation between ##\frac{1}{\cis \theta}## and ##\cis (-\theta)##?
     
  8. Apr 24, 2016 #7
    ##z^{-1}##, but my proof in the OP only works for positive integers. I'm trying to use -1 as a base case to write a proof for negative numbers, but for some reason I cannot work out how to apply a similar strategy (expand then use compound identities)

    It doesn't matter if θ is positive or negative, what matters is its multiplier, which is n.

    Well... they are equal, but thus far my understanding of it is only intuitive, until I can prove it, which is what I'm asking for help with
     
  9. Apr 24, 2016 #8

    Samy_A

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    Ok, so if ##z=a+ib##, where ##a, b## are real, how would you compute ##\frac{1}{z}=\frac{1}{a+ib}##?
    Hint: multiply numerator and denominator by ##a-ib##.

    Once you have proven that ##\frac{1}{\cis \theta}=\cis (-\theta)##, you can easily transform the formula with a negative ##n## into a formula with the positive ##-n##. And then you can use what you already have proven.
     
  10. Apr 24, 2016 #9
    Brilliant, just what I was looking for

    I completely forgot about looking at the complex numbers in rectangular form instead, cheers!
     
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