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Complex Number Proof.

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Using Euler's relation, prove that any complex number z=x+yi can be written in the form z= re[itex]^{i\theta}[/itex] where r and [itex]\theta[/itex] are real. Describe the significance of r and [itex]\theta[/itex] with reference to the complex plane.

    b) Write z= 3+4i in the form z = re[itex]^{i\theta}[/itex]
    (pretty sure I can get this one if I can get help on the proof.

    2. Relevant equations

    e[itex]^{i\theta}[/itex]= cos[itex]\theta[/itex]+isin[itex]\theta[/itex]


    3. The attempt at a solution

    I tried to prove it, got what it wanted me to get but I feel like I did it wrong because I don't know how to go about doing part b. there's also a part c but I didn't feel the need to put it up here because if someone can just explain to me the proof for these equations I think I should be able to get parts b and c
     
  2. jcsd
  3. Feb 5, 2012 #2
    How do cartesian coordinates relate to polar coordinates?
     
  4. Feb 5, 2012 #3
    x = rcos[itex]\theta[/itex]
    y = rsin[itex]\theta[/itex]

    ...is that all you have to do?

    so that makes sense, but I guess I was wrong about knowing how to do part b then... I don't know how to find r and [itex]\theta[/itex] given z = 3+4i
     
  5. Feb 5, 2012 #4
    yes
    So you have your two equations
    [itex]x=r\ Cos( \theta)[/itex]
    [itex]y=r\ Sin( \theta)[/itex]

    How would you find r in terms of x and y?
     
  6. Feb 5, 2012 #5
    r = [itex]\frac{x}{cos\theta}[/itex]

    r = i[itex]\frac{y}{sin\theta}[/itex]


    ??
     
  7. Feb 5, 2012 #6
    nono, r in terms of x and y does not contain any mention of [itex]\theta[/itex]

    Make use of the fact that [itex]Cos( \theta)^2 + Sin( \theta)^2 = 1[/itex].
    You should end up with pythagoras' theorem.

    To find [itex]\theta[/itex], you can make use of [itex]\frac{ Sin(\theta)}{Cos( \theta)} = Tan(\theta )[/itex]
     
  8. Feb 5, 2012 #7
    ok well i'm lost :/

    can't i just take the fact that x = rcos[itex]\theta[/itex] and y= rsin[itex]\theta[/itex] and plug that into z = x +iy? because that'll give the desired results right?
     
  9. Feb 5, 2012 #8
    oh... then i still don't know how to get part b. k well I'll have to figure the whole Pythagorean theorem thing out then
     
  10. Feb 5, 2012 #9
    You can but that isn't going to help you find r and [itex]\theta[/itex]
    I'll show you how to find r, then I'll let you try and find [itex]\theta[/itex]

    1. I'm going to square both of our equations to get

    [itex]x^2 = r^2 \ Cos(\theta )^2[/itex]
    [itex]y^2 = r^2 \ Sin(\theta )^2[/itex]

    2. Next I'm going to add these equations together

    [itex]x^2 + y^2 = r^2 \ Cos(\theta )^2 + r^2 \ Sin(\theta )^2[/itex]

    3. I'm going to pull out a common factor of [itex]r^2[/itex]

    [itex]x^2 + y^2 = r^2 \ (Cos( \theta )^2 + Sin( \theta)^2 )[/itex]

    4. I now use the fact that [itex]Cos( \theta )^2 + Sin( \theta )^2 = 1[/itex] to find

    [itex]x^2 + y^2 = r^2[/itex]

    5. Taking the square root of both sides

    [itex]\sqrt{x^2 + y^2} = r[/itex]

    Which as I said before gives us pythagoras' theorem


    So in b) you have z = 3 + 4i, we can now find the corresponding r, [itex]r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5[/itex]

    All that's left now is to find [itex]\theta[/itex]
     
  11. Feb 5, 2012 #10
    oh my god! thank you! I didn't know you could just add them together sorry my brain is just refusing to work right now but yeah I see how you can get theta now. thank you so much :)
     
  12. Feb 5, 2012 #11
    No problem buddy!
     
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