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Understanding Polar Coordinates and the exponential function

  1. Feb 23, 2013 #1
    I'm reviewing math material for the EIT exam, I'm going over math concepts that should be pretty basic but I feel like there are gaps in my understanding. I understand how we can use rectangular coordinates and complex numbers to find a point on the complex plane. It would follow logically from trig that the rectangular coordinates

    z=x+iy
    that
    z=r(cos(θ)+isin(θ)) being that x=cos(θ) and y=sin(θ)

    I also know the definition z=re however why is this definition true? Can anyone explain it to me?

    In addition to this it can easily be assumed that since z=re and z=r(cos(θ)+isin(θ)) that e=(cos(θ)+isin(θ)) giving us Euler's formula. My next question comes from exactly that.

    How is it that trig functions relate to the exponential function is this just something to accept or is there more underlying causes? Did Euler come up with his equation from the addition of the Maclaurin series of e, sin, and cos; or did he figure this out some other way?
     
  2. jcsd
  3. Feb 23, 2013 #2

    mfb

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    There is a factor of "r" missing for x and y.

    What do you mean "why is this definition true"? Definitions are true because they are defined to be so, but I don't see any definition here.

    You don't need a specific formula for the complex exponential function, you can prove the Euler formula and the Wikipedia page should give you an idea how.
     
  4. Feb 23, 2013 #3
    Sorry I forgot the r.

    I mean how did someone figure out that z=re? Was that realized first so then it led to Euler's formula e = cos(θ)+isin(θ) or was z=re realized from Euler's forumla?
     
  5. Feb 23, 2013 #4

    mfb

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    a=5
    b=7
    How did I figure this out? I did not. I just defined a and b to be 5 and 7.
    In a similar way, for an arbitrary z, you can define r and θ such that they are real and re=z. The Euler formula allows to prove this.
     
  6. Feb 23, 2013 #5

    lurflurf

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    It is obvious as the defining relation of an exponential function if

    exp(x+y)=exp(x)*exp(y)

    and circular functions satisfy

    cos(a+b)+i*sin(a+b)=(cos(a)+i*sin(a)*(cos(b)+i*sin(b))

    and are therefore exponential.

    Euler did manipulate the Maclaurin series. The important thing is any reasonable definition of the complex exponential will relate to trigonometric functions. Any supposed derivation just reveals that fact.
     
    Last edited: Feb 23, 2013
  7. Feb 24, 2013 #6

    HallsofIvy

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    Use power series. The McLaurin series for ex is
    [tex]e^x= 1+ x+ \frac{x^2}{2}+ \frac{x^3}{3!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot[/tex]

    Now, replace x with ix:
    [tex]e^{ix}= 1+ ix+ \frac{(ix)^2}{2}+ \frac{(ix)^3}{3!}+ \cdot\cdot\cdot+ \frac{(ix)^n}{n!}+ \cdot\cdot\cdot[/tex]
    Now, it is easy to see that i2= -1, i3= -i, i4= 1, etc. s0
    [tex]e^{ix}= 1+ ix- \frac{x^2}{2}- i\frac{x^3}{3!}+ \cdot\cdot\cdot+ i^n\frac{x^n}{n!}[/tex]
    and, separating real and imaginary parts,
    [tex]e^{ix}= \left(1- \frac{x^2}{2}+ \frac{x^4}{4!}+ \cdot\cdot\cdot\right)[/tex][tex]+ i\left(x- \frac{x^3}{3!}+ \cdot\cdot\cdot\right)[/tex]
    which are just the McLaurin series for cos(x) and sin(x):
    [tex]e^{ix}= cos(x)+ i sin(x)[/tex]

    We then have ez= ex+ iy= exeiy= ex(cos(y)+ i sin(y)).

    Of course, the fact that [itex]z= x+ iy= r(cos(\theta)+ i sin(\theta))[/itex] comes immediately from conversion of polar coordinates to Cartesian coordinates: [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex].
     
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