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

Homework Help: Euler's Formula help

  1. Apr 13, 2010 #1
    Theres a couple problems im working on that involve complex numbers or euler's formula.
    e^+-(ix) = cos(x) +- isin(x)

    -----------------------
    1. A complex number can be written in rectangular coordinates as z = x+ jy. Write the relations to calculate the
    polar form, z = (r,theta) or z = re^(j*theta) .

    For this one im more confused about what he's asking or how to show the work... i think
    r = sqrt(x^2 + y^2)
    and
    theta = tan^-1(y/x)

    But i'm not really sure if thats what he's looking for


    -----------------------
    3. Convert cos(wt + f) into the sum of complex exponentials.
    Now i know that cos(x) = (e^(ix) + e^-(ix))/2

    Is this as simple as replacing x with (wt + f)?


    ------------------------
    5. Compute [(1+ i*sqrt(3))/2]^2 and (1 + j)^4
    a) directly (using rectangular representation)
    b) using complex exponentials

    How do i go about this for both of these approaches, im not entirely sure how to do either approach.
     
  2. jcsd
  3. Apr 13, 2010 #2
    For 1 and 3 it seems you are on the right track.

    For 5, a) directly using the rectangular representation means just multiplying it out, and b) using complex exponentials means converting the numbers to their complex exponential form and then raising them to the powers.

    Evidently 5 is intended as an object lesson in how much nicer it is to exponentiate the complex form compared to the (a+bi) form, even though most students who are first encountering Euler's formula are much more comfortable with the more standard algebraic method because it's just grinding coefficients.
     
  4. Apr 13, 2010 #3
    Okay.. so i'm a little stuck on 5b...
    [1 + i * sqrt(3)/2]^2

    would i do:

    r = sqrt[1^2 + (sqrt(3)/2)^2]
    theta = tan^-1[(sqrt(3)/2)/1]

    = [r * e ^ (i * theta)] ^ 2
     
  5. Apr 13, 2010 #4
    I just tried working 5b through for (1+I)^4:

    r = sqrt(1^2 + 1^2) = sqrt(2)
    theta = tan^-1(1/1) = pi/4

    ... so

    (1 + i)^4 = [sqrt(2) * e^(i*pi/4)] ^ 4

    apply euler's...

    = (sqrt(2) * [cos(pi/4) + i * sin(pi/4)])^4

    ..simplify

    = (sqrt(2) * [sqrt(2)/2 + i * sqrt(2)/2])^4

    ..multiply

    = (1 + i)^4

    ..well... duh.

    I think my approach might be wrong.
    --------------------------------------
    EDIT
    figured it out:

    (1 + i)^4 = [sqrt(2) * e^(i*pi/4)] ^ 4
    = sqrt(2)^4 * e^(4i*pi/4)
    = 4 * e^(i*pi)
    = 4*-1
    =-4
     
    Last edited: Apr 13, 2010
  6. Apr 14, 2010 #5
    One last thing

    x1(t) = 5cos(400pi*t +0.5pi)
    x2(t) = 5cos(400pi*t -0.25pi)
    x3(t) = 5cos(400pi*t +0.4pi)
    x4(t) = 5cos(400pi*t - 0.9pi)

    I need to express each of those as complex exponentials.. then express the sum.. the only way i know how to do that is to use the fact that
    cos(x) = [e^(i*x)+ e^-(i*x)]
    and make x = 400pi*t +0.5pi.. etc

    which gives me four really long equations.. which seems like it cant possibly be right for what is being asked. At the very least tehre has to be a better way to express the sum of all of them?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook