1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trigometric addition of complex numbers!

  1. Apr 19, 2007 #1
    Hello

    I got some funny idears don't know if they are true, but I will share them with you.

    1. The problem statement, all variables and given/known data

    I would like to prove these formula
    (1)[tex]sin(z_1 + z_2) = sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1)[/tex]

    (2)[tex]cos(z_1 + z_2) = sin(z_1) \cdot sin(z_2) - cos(z_2) \cdot cos(z_1)[/tex]

    (3)[tex]cos(z)^2 + sin(z)^2 = 1[/tex]

    for any complex number [tex]z_1, z_2, z \in \mathbb{C}[/tex]

    3. The attempt at a solution
    Proof for (1)

    (step a)Let [tex]z_1 = x_1 + i \cdot y_1[/tex] and [tex]z_2 = x_2 + i \cdot y_2[/tex]

    (step b)According to the rules of addition of complex numbers

    [tex]z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) [/tex], where [tex]z_1 + z_2 \in \mathbb{C}[/tex]

    (step c)I denote [tex]x_1 + x_2 = x [/tex] and [tex]y_1 + y_2 = y[/tex]

    (step d)thus by replacing x and y by their respective polar coordinant representation:

    [tex]z_1 + z_2 = (r \cdot cos(\theta)) + i \cdot ( r \cdot sin(\theta)) [/tex], where [tex]z_1 + z_2 \in \mathbb{C}[/tex]

    (step e)which can be reformulated to:

    [tex]z_1 + z_2 = r \cdot (cos(\theta)) + i \cdot ( sin(\theta)) [/tex].

    (step f)Which equals according to the euler identity:

    [tex]z_1 + z_2 = r \cdot e^{i \theta} [/tex].

    (step g)Then I apply sinus on both sides of the equality and get

    [tex]\begin{array}{ccc}sin(z_1 + z_2) = cosh(sin(\theta) \cdot r) \cdot sin(cos(\theta) \cdot r) + sinh(sin(\theta) \cdot r) \cdot cos(cos(\theta) \cdot r) \cdot i \end{array}[/tex]

    From the above we see that [tex]x = r \cdot cos(\theta)[/tex] and [tex]y = r \cdot sin(\theta)[/tex]

    (step h)
    With this in mind we rearrange the above and remembers the identity for x and y.

    [tex]sin(z_1 + z_2) = sin(x_1 + x_2) \cdot cosh(y_1 + y_2) + cos(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i[/tex]

    I have thus proved the sinus addition formula for complex numbers. (q.e.d)

    (2) Proof

    The steps are the same until (step g) where I take cosine on both sides of the equality and then obtain

    [tex]cos(z_1 + z_2) = cos(x_1 + x_2) \cdot cosh(y_1 + y_2) - sin(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i[/tex]

    q.e.d.

    Test for (1): Let [tex]z_1 = 2 + 2i[/tex] and [tex]z_2 = 4 + 2i[/tex]

    Then by applying my formula on

    [tex]z_1 + z_2 = 6 + 4i[/tex]

    [tex]sin(z_1 + z_2) = sin(6 + 4i) = sin(6) \cdot cosh(4) + cos(6) \cdot sinh(4) \cdot i[/tex]

    Thus my formula is true.

    Test for (1): Let [tex]z_1 = 2 + 2i[/tex] and [tex]z_2 = 4 + 2i[/tex]

    Then by applying my formula on

    [tex]z_1 + z_2 = 6 + 4i[/tex]

    [tex]cos(z_1 + z_2) = cos(6 + 4i) = cos(6) \cdot cosh(4) - sin(6) \cdot sinh(4) \cdot i[/tex]

    Thus my formula is true.


    Best Regards

    Fred

    2. Relevant equations


    p.s. Does my proof look okay? If yes could somebody please give a hint for (3)??
     
  2. jcsd
  3. Apr 19, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    I haven't looked through the 2 proofs very well, but they look fine. For 3, Write sin x and cos x in exponential form from Eulers Formula. Write z = a+bi, and expand.
     
  4. Apr 19, 2007 #3
    Hi Gib and Dick,

    Thank You for Your answer,

    Regardin (3)

    Proof:

    Let [tex]z = a +ib[/tex] be a complex number.

    Then according to Eulers formula:

    [tex]sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/tex]

    [tex]cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/tex]

    Then by squaring the above and adding them together (My motovation for doing this part, is that that the point (sin(z),cos(z)) lies on the unit cicle (x^2 + y^2 = 1) ?)

    [tex](\frac{e^{iz} - e^{-iz}}{2i})^2 + (\frac{e^{iz} + e^{-iz}}{2})^2 =1[/tex]

    Thus the formula for [tex]sin(z)^2 + cos(z)^2 =1[/tex]

    where z is complex number is true. (q.e.d)

    Best Regards
    Fred

    p.s. You you please be so kind to comment my proofs for (1) and (2) :)

    p.p.s. Thank You Dick for you correction of my trig.add mistakes. It was a dumb a** mistake on my part.
     
    Last edited: Apr 19, 2007
  5. Apr 19, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Square the exponential expressions and add them. cos(z) equals negative cos(a+ib)!? Don't be silly.
     
  6. Apr 19, 2007 #5
    Hi Dick,

    Please look at my edited post #3.

    Best Regards
    Fred
     
  7. Apr 19, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks fine. Did you actually expand the squares and use the rules of exponentials to verify the sum is one?
     
  8. Apr 19, 2007 #7
    No I didn't.

    I will just look it up.
     
    Last edited: Apr 19, 2007
  9. Apr 19, 2007 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    NO. Like this:

    [tex] (\frac{e^{iz} + e^{-iz}}{2})^2+(\frac{e^{iz} - e^{-iz}}{2i})^2 = ?? = 1[/tex]

    Where did that crazy sum come from?
     
    Last edited: Apr 19, 2007
  10. Apr 19, 2007 #9
    Dear Dick

    It was a cut and past error on my part , it should have said

    (sin(z))^2 + (cos(z))^2 = 1.

    Best regards

    Fred

    p.s. You mean by expand the above I write them as the sum of their squared respective power series?
     
    Last edited: Apr 19, 2007
  11. Apr 19, 2007 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No again. (cos(z))^2=((exp(iz)+exp(-iz))/2)^2=exp(2iz)/4+1/2+exp(-2iz)/4. Do you see why? What's sin(z)^2? Add them.
     
  12. Apr 19, 2007 #11
    Hello Dick,

    I forgot to be carefull. Sorry.

    By repeating you thema I get sin(z)^2


    [tex]sin(z)^2 = (\frac{e^{iz}-e^{-iz}}{2i})^2 = \frac{-(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}[/tex]

    Therefore sin(z)^2 + cos(z)^2 = 1.

    Best Regards
    Fred
     
    Last edited: Apr 19, 2007
  13. Apr 19, 2007 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    And adding them gives you?
     
  14. Apr 19, 2007 #13
    I changed my post #11 above.

    How does the proof of (3) look now? Am I steppin in the right direction?

    Best Regards.

    /Fred
     
    Last edited: Apr 19, 2007
  15. Apr 19, 2007 #14

    Gib Z

    User Avatar
    Homework Helper

    Looks like the proof is complete. Well done, I guess there was no need to let z=a+bi after all.
     
  16. Apr 20, 2007 #15
    Hi Just a clean write of part 3,

    Proof (Part 3):

    Let [tex]z[/tex] be a complex number.

    Then according to Eulers formula:

    [tex]sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/tex]

    [tex]cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/tex]


    Its know that

    [tex](sin(z))^2 = (\frac{e^{iz} - e^{-iz}}{2i})^2 = -\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4} [/tex]

    [tex](cos(z))^2 = (\frac{e^{iz} + e^{-iz}}{2})^2 = \frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4} [/tex]

    By addin them together I get

    [tex](sin(z))^2 + (cos(z))^2 = (-\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}) + (\frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4}) = 1 [/tex]

    Thus the formula for [tex]sin(z)^2 + cos(z)^2 = 1[/tex]

    where z is complex number is true. (q.e.d)

    Best Regards
    Fred

    p.s. Could somebody please be so kind and look through this part 3, and see if I have overlooked anything?
     
    Last edited: Apr 20, 2007
  17. Apr 20, 2007 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks good to me.
     
  18. Apr 21, 2007 #17
    Regarding (1) and (2) are they okay too?

    Any need to try to factor my formula result so they look like the original formulas

    sin(z1 + z2) = sin(z1) * cos(z2) + sin(z2) * cos(z1) ?

    or is my way of formulating it consistant??

    Best Regards
    Fred
     
  19. Apr 21, 2007 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    (1) and (2) don't look so great. Just do them the same way you did (3). Replace all of the cos's and sin's by exponentials on both sides of the equations. Then multiply everything out and show both sides are equal. BTW you have the two terms on the RHS reversed in your statement of (2).
     
  20. Apr 21, 2007 #19
    Okay thanks,

    Thought I was too lucky about 1 and 2.

    by using the euler formula for sin(z1+z2)

    I get following:

    [tex]\begin{array}{cccc}sin(z_1+z_2) &=& \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \cdot \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} \\ &=& sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1) \end{array} [/tex]


    By using euler on cos(z_1 + z_2)

    [tex]\begin{array}{cccc}cos(z_1+z_2) &=& \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \\ &=& cos(z_1) \cdot cos(z_2) - sin(z_1) \cdot sin(z_2) \end{array} [/tex]


    Is that what you mean?

    Best regards

    Fred
     
    Last edited: Apr 21, 2007
  21. Apr 21, 2007 #20

    Gib Z

    User Avatar
    Homework Helper

    Yes that Looks excellent. Good work.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trigometric addition of complex numbers!
  1. Complex numbers (Replies: 36)

Loading...