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Complex numbers

  1. Nov 21, 2006 #1
    [tex] Re^{j \theta} = R{(cos \theta + jsin\theta )}[/tex]

    can anyone show me this proof or show me a link please
     
    Last edited: Nov 21, 2006
  2. jcsd
  3. Nov 21, 2006 #2

    dextercioby

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    What do you know about Taylor series ?

    Daniel.
     
  4. Nov 21, 2006 #3
    I haven't come across this Taylor series much, using this I take it the proof is harder that i suspected
     
  5. Nov 21, 2006 #4

    Hootenanny

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    You can also prove this result using calculus. If you have met differentiation, try taking the derivative of f(x) as see where it takes you,

    [tex]f(x):=\frac{R\cos(x)+iR\sin(x)}{Re^{ix}}\hspace{1cm}x\in\Re\;\;\; , \;\;\; R\neq0[/tex]
     
    Last edited: Nov 21, 2006
  6. Nov 21, 2006 #5
    Although my consistency with things this level arent good here goes

    [tex]f(x)= \frac {Rcosx + jRsinx} {Re^{jx}}

    = \frac {ln(Rcosx) + ln(jRsinx)} {ln(Re^{jx}}

    = \frac {-Rsinx} {Rcosx} + \frac {jRcosx} {jRsinx} /j

    = \frac {jcotx - tanx} {j}[/tex]
     
  7. Nov 21, 2006 #6

    Hootenanny

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    I'm afraid your technique is not quite correct. Have you met the Quotient Rule yet? Also, I have noticed that you are using [itex]j[/itex] in your questions; I am assuming here you mean the imaginary unit [itex]i=\sqrt{-1}[/itex]. It may also perhaps be desirable to remove the [itex]R[/itex] term for the moment to remove the condition that R be non zero. Thus, we have;

    [tex]f(x):=\frac{\cos(x)+i\sin(x)}{e^{ix}}\hspace{1cm} x\in\Re[/tex]
     
    Last edited: Nov 21, 2006
  8. Nov 21, 2006 #7
    [tex]

    f(x) =\frac {e^{ix}(icosx-sinx) - ie^{ix}(cosx + isinx)} {(e^{ix})^2}

    f(x) =\frac {ie^{ix}cosx - e^{ix}sinx - ie^{ix}cosx - i^2e^{ix}sinx} {(e^{ix})^2}

    f(x) =\frac {e^{ix}sinx + e^{ix}sinx} {(e^{ix})^2}

    f(x) =\frac {2sinx} {e^{ix}}
    [/tex]
     
    Last edited: Nov 22, 2006
  9. Nov 21, 2006 #8

    Hootenanny

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    Your getting closer, I'll show you;

    [tex]\frac{d}{dx}\left (\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]

    So in this case [itex]u=\cos(x) + i\sin(x)[/itex] and [itex]v=e^{ix}[/itex], thus;

    [tex]\frac{du}{dx} = -\sin(x) + i\cos(x)\hspace{2cm}\frac{dv}{dx} = i\cdot e^{ix}[/tex]

    Thus we obtain;

    [tex]f'(x) = \frac{e^{ix}\cdot (-\sin(x) + i\cos(x)) - (\cos(x) + i\sin(x))\cdot i\cdot e^{ix}}{(e^{ix})^2}[/tex]

    [tex]=\frac{-e^{ix}\cdot\sin(x) + e^{ix}\cdot i\cos(x) - e^{ix}\cdot i \cdot\cos(x) - e^{ix}\cdot i^2\cdot\sin(x)}{(e^{ix})^2}[/tex]

    Noting that the two consine terms cancel and that [itex]i^2=(\sqrt{-1})^2 = -1[/itex] ;

    [tex]f'(x) = \frac{-e^{ix}\cdot\sin(x) + e^{ix}\cdot\sin(x)}{(e^{ix})^2}[/tex]

    [tex]= \frac{-\sin(x)+\sin(x)}{e^{ix}} = \frac{0}{e^{ix}} = 0[/tex]

    What can we say about a function if it has a derivative of zero everywhere in its domain?
     
  10. Nov 22, 2006 #9
    I see now, silly mistakes by me.
    When a derivative is zero then the things differentiated must have been a constant
     
  11. Nov 22, 2006 #10
    From a book i have read in complex analysis, and from what i understand about the subject, I believe that we just define this equation. I mean, we define that [tex] e^{a+j \theta} = e^a{(cos \theta + jsin\theta )}[/tex] . It's simply the definition for complex exponents. You can also reach this equation by using the Taylor series of e^x=1+x+(x^2)/2!+... , using x=ja but it's not any real proof unless you prove that the series converges for any imaginary number ja. Also, how can you check the above equation while the left part of it cannot be interpreted in a certain way?? So, i see it as a definition and not as an equation that you can prove.
     
    Last edited: Nov 22, 2006
  12. Nov 22, 2006 #11
  13. Nov 22, 2006 #12

    robphy

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    How about this?

    Write [tex]e^{j\theta}=E(\theta)+jO(\theta) [/tex],
    where [tex]E(\theta)=\frac{1}{2} (e^{j\theta}+e^{-j\theta})[/tex]
    and [tex]O(\theta)=\frac{1}{2j} (e^{j\theta}-e^{-j\theta})[/tex],
    where the right-hand side is simply the sum of the even and odd parts of the left-hand side. [tex]j\neq 0[/tex] has no particular meaning right now, except that it is constant (independent of [tex]\theta [/tex]).

    Then [tex]
    \begin{align*}
    \frac{d}{d\theta} e^{j\theta} &= \frac{1}{2}\left( j (e^{j\theta}-e^{-j\theta} \right)+
    j\frac{1}{2j}\left( j (e^{j\theta}+e^{-j\theta} \right)\\
    je^{j\theta} &= j^2O(\theta)+jE(\theta)
    \end{align*}
    [/tex]
    where we have noted that
    [tex]\frac{d}{d\theta}E(\theta)=j^2O(\theta)[/tex] and [tex]\frac{d}{d\theta}O(\theta)=E(\theta)[/tex]... a coupled set of differential equations.

    Continuing on, we find [tex]\frac{d^2}{d\theta^2}E(\theta)=j^2(E(\theta))[/tex] and [tex]\frac{d^2}{d\theta^2}O(\theta)=(j^2O(\theta))[/tex]... two differential equations... or simply, two questions: what functions are proportional to their second derivatives? Let's now formally require that [tex]j^2=-1[/tex]. So, now our question is: what functions are equal to minus their second derivatives? (With constants of integration, you'll need to consider the initial conditions to completely determine the functions E and O.)

    I'll stop here... but you should be able to finish this off.
     
    Last edited: Nov 22, 2006
  14. Nov 22, 2006 #13
    We want to show,

    [tex] e^{ix} = \cos x + i \sin x [/tex]

    Taylor expanding yields:

    [tex] \sin x = x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots [/tex]
    [tex] \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots [/tex]

    [tex] e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \ldots [/tex]

    Finish this off. Remember [itex] (i)^2 = -1 [/itex]

    ex) [tex] (ix)^5 = i^2 i^2 i x^5 = (-1)(-1)i x^5 = ix^5 [/tex]
     
    Last edited: Nov 22, 2006
  15. Nov 23, 2006 #14

    Hootenanny

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    Indeed, that is correct. So, try taking our function, let [itex]x=0[/itex] and see where this takes you.

    Of course Euler's formula can be derived using Taylor's series (as dextercioby and Frogpad said), you could also prove this using differentials; I was showing this method simply because you said that you hadn't met Taylor series before, but all proofs are equally valid,although I myself find the Taylor series most elegant.
     
  16. Nov 24, 2006 #15
    well my original question has been answered now so thank you all.
    as for that last this hootenanny do you mean e^ix= or f(x)=

    f(x)=(1+0)/1=1 and e^ix: 1=1+0
    but i dont understand what either of these prove
     
  17. Nov 24, 2006 #16

    Hootenanny

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    We have;

    [tex]f(x):=\frac{\cos(x)+i\sin(x)}{e^{ix}}[/tex]

    Since we know that f(x) is a constant function it has the same value everywhere in its domain. Therefore, we can say that;

    [tex]f(x) = f(0) \;\;\; \forall \;\;\;x \Rightarrow \frac{\cos(x)+i\sin(x)}{e^{ix}} = \frac{\cos(0)+i\sin(0)}{e^{i0}} = \frac{1}{1}[/tex]

    [tex]\therefore \frac{\cos(x)+i\sin(x)}{e^{ix}} = 1 \Leftrightarrow \cos(x)+i\sin(x) = e^{ix}[/tex]

    [tex]\Rightarrow R(\cos(x)+i\sin(x)) = R\cdot e^{ix}\hspace{1cm}\text{Q.E.D.}[/tex]

    Hope this was helpful.
     
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