1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig question

  1. Jan 17, 2009 #1
    [tex] | a + bi | = \sqrt{(a)^2+(b)^2} [/tex]
    [tex] \theta = tan^{-1}(\frac{b}{a}) [/tex]

    [tex] |(a + bi)^n| = (\sqrt{(a)^2+(b)^2})^n[/tex]
    [tex] \theta = n(tan^{-1}(\frac{b}{a}))[/tex]

    [tex] (a+bi) = (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

    Why is [tex]\theta = n(tan^{-1}(\frac{b}{a}) [/tex] ? For example when I have

    [tex]5 + i[/tex]

    [tex]\theta = tan^{-1}(\frac{1}{5}) = 0.1974...[/tex]

    [tex](5+i)^2 = 25 + 10i -1[/tex]

    [tex](5+i)^2 = 24 + 10i[/tex]

    [tex]\alpha = tan^{-1}(\frac{10}{24}) = 0.3948...[/tex]

    [tex]2(0.1974) =0.3948[/tex]

    Why is it that the exponent of the vector can be used to get the angle of the resultant by simply multiplying it with the tan function? Also in the first part of that:

    [tex] (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

    Just to make sure, it is only

    [tex] isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

    and not simply

    [tex] (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + sin(n(tan^{-1}(\frac{b}{a}))))[/tex]

    because one of the components is complex?
  2. jcsd
  3. Jan 17, 2009 #2


    User Avatar
    Science Advisor

    you are talk about putting complex numbers in "polar form". If you represent a+ bi as the point (a, b) in Cartesian coordinates and the polar form is [itex](r, \theta)[/itex] (That is, distance from (0,0) to (a,b) is r and the line from (0,0) to (a,b) is [itex]\theta[/itex], it is easy to see that [itex]a= rcos(\theta)[/itex] and [itex]b= rsin(\theta)[/itex] so that [itex]a+ bi= r (cos(\theta)+ i sin(\theta))[/itex]. Then [/itex](a+bi)^2= r^2(cos^2(\theta)-sin^2(\theta)+ i(2sin(\theta)cos(\theta)))[/itex] and recognise that [itex]cos^2(\theta)- sin^2(\theta)= cos(2\theta)[/itex] and [/itex]2cos(\theta)sin(\theta)= sin(2\theta)[/itex]. The more general formula follows from the identites for [itex]sin(\theta+ \phi)[/itex] and [itex]cos(\theta+ \phi)[/itex].

    Even simpler is to note that [itex]cos(\theta)+ i sin(\theta)= e^{i\theta}[/itex] so that [itex](cos(\theta)+ i sin(\theta)^n= (e^{i\theta})^n= e^{ni\theta}[/itex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Trig question
  1. Trig question (Replies: 2)

  2. Trig question (Replies: 8)

  3. Trig question (Replies: 5)