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Complex Numbers

  1. Oct 9, 2007 #1
    1) A mathemetician is willing to sell you something valued at $i^i. Would you pay him 20 cents for it?

    2) Let z=(z1/z2) where z1 = a+ib and z2 = c+id. Show the angle of z is the difference between angle z1 and z2.

    3) Show that multiplying any vector by e^ix doesn't alter its length.

    My attempts:

    1) Tried using cosx + isinx to no avail. Honestly have no idea where to begin.
    2) First I found the z by multiplying it out. let x be the angle. so i did tanx = bc-ad/ac+bd. Individually, you get tanx1 = b/a and tanx2=d/c which is not the same.
    3) in the length formula i end up with cos^2-sin^2 which does alter length.

    ruined my thanks giving, so i give up onthese.
  2. jcsd
  3. Oct 9, 2007 #2
    You're more or less on the right track...use Euler's formula/identity.

    Again, this is correct. You just need to show that difference of the angles is equal to arctan((bc-ad)/(ac+bd)). Look up trig. identites.

    I don't quite understand what question 3 asks for. Sorry.
  4. Oct 13, 2007 #3
    Euler's Identity

    So you're solving for [tex]i^i[/tex]. The Euler's Identity states that [tex]e^{\pi i}+1=0[/tex]. Using just this identity, many other quantities can be derived, including our [tex]i^i[/tex].

    Start by subtracting [tex]1[/tex] from both sides, then getting the square root of both sides:
    [tex]e^{\frac{1}{2}\pi i}=(-1)^{\frac{1}{2}}[/tex]

    [tex](-1)^{\frac{1}{2}}[/tex] obviously can be written as [tex]\sqrt{-1}[/tex], which simplifies to [tex]i[/tex]:
    [tex]e^{\frac{1}{2}\pi i}=i[/tex]

    Now raise both sides to the power of [tex]\frac{1}{i}[/tex]:
    [tex]e^{\frac{1}{2}\pi i\frac{1}{i}}=i^{\frac{1}{i}}[/tex]

    Since [tex]\frac{1}{i}[/tex] equals [tex]i^{-1}[/tex], substitute:

    Now simply raise both sides to the power of [tex]-1[/tex]:

    And voila: [tex]i^{i} = e^{-\frac{1}{2}\pi} = 0.207879576 ...[/tex]

    Now you be the economist and tell me whether you will accept the offer or not.
  5. Oct 13, 2007 #4
    Close. We want to have the magnitude (absolute value) of e^ix = 1. So we write e^ix in the form of cos(x)+isin(x). Finding the magnitude of this is done by finding the square root of the sum of the squares so sqrt(cos(x)^2 + sin(x)^2). You'll notice that the i dropped out. This is because the definition of the magnitude of a set of terms is simply sqrt(t1^2+t2^2+...tn^2). We don't care that i^2 is -1, we still sum the term.

    I think you'll agree that cos(x)^2+sin(x)^2 is 1 and that sqrt(1) is 1.

    To the mods, the only reason I gave the full answer was that the OP was basically there, s/he just needed to see why it was positive sin(x)^2.
  6. Oct 14, 2007 #5

    Gib Z

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    Homework Helper

    For 3, the simplest way to do it is to write the general complex number in polar form:

    [tex]z= r e^{i\theta}[/tex].

    Which quantity gives the length? Does multiplying by [itex]e^{ix}[/itex] affect this quantity?
  7. Oct 14, 2007 #6
    for second question a quicker way is to convert to the polar form and then u can see the angle of z1/z2

    for first question i=e^i*pi/2 (since sine(pi/2)=1) raise this to "i" power u immediately get e^-pi/2.
    Last edited: Oct 14, 2007
  8. Oct 14, 2007 #7

    Gib Z

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    Homework Helper

    Good work, but try to stick to one notation for the imaginary unit at a time. Are you a physics or engineering student?
  9. Oct 14, 2007 #8
    lol sorry EE undergraduate hehe

    EDIT: everything changed to i now
  10. Oct 14, 2007 #9
    That's not really a proof. That's saying that e^xi doesn't affect the length because... well it doesn't affect the length.
  11. Oct 14, 2007 #10


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    Staff Emeritus
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    Gold Member

    I disagree. You can always calculate the norm of the vector before and after, and show that it is the same.
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