Results 1 to 5 of 5

Thread: Complex Analysis argument question

  1. #1
    Junior Member
    Joined
    Oct 2007
    From
    Nova Scotia
    Posts
    48

    Complex Analysis argument question

    Could someone help me on this one.
    Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by frankdent1 View Post
    Could someone help me on this one.
    Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.
    If $\displaystyle z\not = 0$ then write $\displaystyle z = |z|e^{\theta i}$ where $\displaystyle \theta = \arg (z)$. This means $\displaystyle z^n = |z^n|e^{n\theta i}$ and that means $\displaystyle n\theta $ is the $\displaystyle \arg(z^n)$ (up to a factor of $\displaystyle 2\pi$).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2007
    From
    Nova Scotia
    Posts
    48
    Would you care to elaborate on the last part?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by frankdent1 View Post
    Would you care to elaborate on the last part?
    I am saying if $\displaystyle \theta$ is the argument of a non-zero complex number then we can write $\displaystyle z = |z|e^{i\theta}$ (and conversely). Now if we raise both sides to the $\displaystyle n$ we get $\displaystyle z^n = |z|^n \left( e^{i\theta} \right)^n$ but $\displaystyle |z|^n = |z^n|$ and $\displaystyle \left( e^{i\theta} \right)^n = e^{ni\theta}$ by de Moivre theorem. Thus, we can write $\displaystyle z^n$ as $\displaystyle |z^n|e^{ni\theta}$ and so $\displaystyle ni\theta$ is the argument of $\displaystyle z^n$, i.e. $\displaystyle n\theta = \arg (z^n)$ thus $\displaystyle n\arg (z) = \arg (z^n)$. I said "up to a factor of $\displaystyle 2\pi$" because as you know there can be many different argument differening by $\displaystyle 2\pi$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by ThePerfectHacker View Post
    If $\displaystyle z\not = 0$ then write $\displaystyle z = |z|e^{\theta i}$ where $\displaystyle \theta = \arg (z)$. This means $\displaystyle z^n = |z^n|e^{n\theta i}$ and that means $\displaystyle n\theta $ is the $\displaystyle \arg(z^n)$ (up to a factor of $\displaystyle 2\pi$).
    The red word factor is probably not the word I'd use since it implies multiplication of the multiples of 2 pi, whereas you are adding the multiples of 2 pi.

    Quote Originally Posted by frankdent1 View Post
    Would you care to elaborate on the last part?
    If you think of z^n as representing some complex number (1 + i, say) and then think of an Argand diagram showing 1 + i, you should be able to that adding (or subtracting) multiples of 2 pi 'keeps you' at 1 + i. Capisce?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex analysis T/F question
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: Nov 23rd 2011, 03:45 PM
  2. Replies: 3
    Last Post: Oct 4th 2011, 05:30 AM
  3. Complex analysis question...
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 27th 2010, 05:29 AM
  4. Argument Principle, Complex Analysis
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 25th 2010, 01:30 PM
  5. Replies: 6
    Last Post: Jun 20th 2009, 07:40 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum