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Homework Help: Complex Numbers 2

  1. Sep 26, 2007 #1
    Consider the complex number z=e[tex]_{i\theta}[/tex] = cos[tex]\theta[/tex]+isin[tex]\theta.[/tex] By evaluating z[tex]^{2}[/tex] two different ways, prove the trig identities cos2[tex]\theta[/tex] = cos[tex]^{2}[/tex][tex]\theta[/tex] - sin[tex]^{2}[/tex][tex]\theta[/tex] and sin2[tex]\theta[/tex] = 2sin[tex]\theta[/tex]cos[tex]\theta[/tex].

    A question about the approach to this question:
    How do you guys approach the task of 'evaluating' something, when told to do so, like here.
    I find myself doing random manipulations without knowledge of whether the road I'm on is even close to the correct path or not.
    Evaluate seems like such a general instruction...

    If I square z I get;
    z[tex]^{2}[/tex] = cos[tex]\theta[/tex][tex]^{2}[/tex] + i[tex]^{2}[/tex]sin[tex]\theta[/tex][tex]^{2}[/tex]

    z[tex]^{2}[/tex] = cos[tex]\theta[/tex][tex]^{2}[/tex] - sin[tex]\theta[/tex][tex]^{2}[/tex]

    If I sub z[tex]^{2}[/tex] back in I get 0 so that's wrong.

    Is the use of the 'e definition' necessary for this? I kind of want to see how I ought to be approaching this 'evaluation'.

    Thanks, as always
  2. jcsd
  3. Sep 26, 2007 #2

    If [tex]z = \cos{\theta}+\imath\sin{\theta}[/tex], then [tex]z^2 = \left(\cos{\theta}+\imath\sin{\theta}\right)^2 \neq \cos^2{\theta}+\imath^2\sin^2{\theta}[/tex].

    Square that. Then use the other way of evaluating a complex number raised to some power, i.e., de Moiver's formula. Equate the two.
    Last edited: Sep 26, 2007
  4. Sep 26, 2007 #3
    Am I allowed to do this?


    =-sin[tex]\theta[/tex][tex]^{2}[/tex] ?
  5. Sep 26, 2007 #4
    The way you have written is quite ambiguous. But if you mean [tex]\left(\imath\sin{\theta}\right)^2[/tex], then it is equal to [tex]-\sin^2{\theta}[/tex].

    P.S. You need not enclose each term of LaTeX code within 'tex' tags. It is enough if you do it for each line of code. Example: Enclosing this - \left(\imath\sin{\theta}\right)^2 - within tex tags will produce [tex]\left(\imath\sin{\theta}\right)^2[/tex]. :smile:
  6. Sep 26, 2007 #5
    lol thanks!
    It just does that when I click the latex icon each time.. your saying you write it all out manually?
  7. Sep 26, 2007 #6
    I'm still stuck with:

    z[tex]^{2}[/tex] = cos[tex]^{2}[/tex][tex]\theta[/tex] + 2isin[tex]^{2}[/tex][tex]\theta[/tex] - sin[tex]^{2}[/tex][tex]\theta[/tex]

    You'll see I still did it the long way if you quote:P
  8. Sep 26, 2007 #7
    While I mostly do it manually(click n' latex is only a recent feature), I'm not asking you to do it manually.

    When I click on any icon, what I get is that piece of code (say, \frac{}{} for fraction) within tex codes. The cursor is still within the those tags, so when I click on another icon, only its code is printed within the same set of tags. When I want to do a new line, I move the cursor outside the first set of tex tags.
  9. Sep 26, 2007 #8
    Shouldn't it be [tex]\cos^2{\theta} + 2\imath\cos{\theta}\sin{\theta} - \sin^2{\theta}[/tex]?
  10. Sep 26, 2007 #9

    yeah, sorry. Typo
  11. Sep 26, 2007 #10
    Did you also find the other expression using de Moivre's theorem/formula?
  12. Sep 26, 2007 #11
    oh i equate the 2 just for one of the two im looking for? I thought that was for the other equation
  13. Sep 26, 2007 #12
    What do you get when you equate the two?
  14. Sep 26, 2007 #13
    I've gotten to

  15. Sep 26, 2007 #14
    im realizing now thats wrong isnt it..
  16. Sep 26, 2007 #15
    When is a + ib = c + id? (a,b,c,d are real)
  17. Sep 26, 2007 #16
    a=c, b=d?
  18. Sep 26, 2007 #17

    So when you equate cos(2t) + isin(2t) and cos^2(t) + 2isin(t)cos(t) - sin^2(t)...?

    (using 't' instead of theta.)
  19. Sep 26, 2007 #18
    well if i use what im trying to prove,
    cos(2t) = cos^2(t)-sin^2(t)
    and sin(2t) = 2sintcost
  20. Sep 26, 2007 #19
    but doesnt this method require that which im proving, to prove it?
  21. Sep 26, 2007 #20
    No. You're just equating the real and imaginary parts of the complex numbers. All you're using is the definition of equality of two complex numbers. (a = c, b = d)
  22. Sep 26, 2007 #21

    So is it a valid proof to say that

    cos2t +isin2t = cos^2(t) + 2isintcost - sin^2(t)

    cos2t must = cos^2(t)-sin^2(t)
    sin2t=2sintcost ?
  23. Sep 26, 2007 #22
    They must be equal since the corresponding real and imaginary components are equal.

    I think writing the RHS as (cos^2(t) - sin^2(t)) + i(2sin(t)cos(t)) will make things familiar to you.
  24. Sep 26, 2007 #23

    If the i's cancel leaving:
    cos2t + sin2t = cos^2(t) - sin^2(t) + 2sin(t)cos(t)

    There are all real parts. You dont think I'll need to prove the two RHS pieces?
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