Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How exp(jwt)=cos(wt)+jsin(wt)?

  1. Feb 19, 2014 #1
    Can anyone explain in simple terms on how they are equal? graphs are welcome.

    The thing I don't understand is why exp(jwt) is sinusoidal. Wouldn't exp(jwt) go to infinity when t increases? I don't know how to understand the exp term as sinusoidal? in my mind they simply go to zero or infinity.
     
  2. jcsd
  3. Feb 19, 2014 #2

    jgens

    User Avatar
    Gold Member

    Assuming j is the imaginary unit just compare the Taylor series for these two functions.
     
  4. Feb 19, 2014 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    exp(x) behaves the way you expect. The image of exp(ix) is the unit circle on the complex plane.
     
  5. Feb 20, 2014 #4

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    notice they solve the same differential equation.
     
  6. Feb 21, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I like Mathwonk's answer! A similar way to see that [itex]e^{ix}= cos(x)+ isin(x)[/itex] is to use the Taylor's series at x= 0 (as jgens said):

    [tex]e^{x}= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot[/tex]
    [tex]cos(x)= 1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n!)+ \cdot\cdot\cdot[/tex]
    [tex]sin(x)= x- x^3/3!+ x^5/5!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot[/tex]

    Each involves powers of x over the factorial of that power. Sine has only odd powers, cosine only even powers and sine and cosine have alternating sign.

    Now see what happens when we replace "x" with "ix" in the series for [itex]e^x[/itex]:
    [tex]e^{ix}= 1+ (ix)+ (ix)^2/2+ (ix)^3/3!+ \cdot\cdot\cdot+ (ix)^n/n!+ \cdot\cdot\cdot[/tex]
    [tex]e^{ix}= 1+ ix+ i^2x^2/2+ i^3x^3/3!+ \cdot\cdot\cdot+ i^nx^n/n!+ \cdot\cdot\cdot[/tex]
    Of course, [itex]i^2= -1[/itex] so [itex]i^3= i^2(x)= -i[/itex], [itex]i^4= -i(i)= 1[/itex] and then it all repeats. That is, every odd power of ix is plus or minus ix while every even power is plus or minus 1. So
    [tex]e^{ix}= 1+ ix- x^2/2- ix^3/3!+ \cdot\cdot\cdot+ (-1)^nx^{2n}/(2n)!+ (-1)^ni x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot[/tex]

    and separating "real" and "imaginary" parts
    [tex]e^{ix}= (1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n)!+ \cdot\cdot\cdot)+ i(x- x^3/3!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!)[/tex]
    [tex]e^{ix}= cos(x)+ isin(x)[/tex]

    That's because you are used to dealing with real numbers. The exponential is increasing along along the real axis, periodic along the imaginary axis.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook