Results 1 to 3 of 3

Math Help - calculation method??

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    China
    Posts
    4

    Question calculation method??

    Dear all
    please help me to understand the calculation method,about the eq(see as follows)

    eq url:
    http://i.imgur.com/1SVsrWr.jpg

    than k u for ur help.

    *************
    Mattew
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,348
    Thanks
    901

    Re: calculation method??

    $\Large x + \imath y = \rho e^{\imath \theta}$

    where
    $\large \rho = \sqrt{x^2 + y^2}$
    $\large \theta = \arg(x,y) = \arctan(y,x)$ where $\arctan(y,x)=\arctan\left(\frac y x\right)$ with the result adjusted for the correct quadrant.

    Then $\Large \left(\rho e^{\imath \theta}\right)^k = \rho^k e^{\imath k \theta}$

    Also if you have two complex numbers in polar form their product is given by

    $\Large \left(\rho_1 e^{\imath \theta_1}\right)\left(\rho_2 e^{\imath \theta_2}\right)=\rho_1 \rho_2 e^{\imath(\theta_1+\theta_2)}$

    that should be all you need to do your calculation
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2013
    From
    China
    Posts
    4

    Re: calculation method??

    Very clear !!!!
    thank u for ur clear answer.

    ************
    Matthew ^^
    Quote Originally Posted by romsek View Post
    $\Large x + \imath y = \rho e^{\imath \theta}$

    where
    $\large \rho = \sqrt{x^2 + y^2}$
    $\large \theta = \arg(x,y) = \arctan(y,x)$ where $\arctan(y,x)=\arctan\left(\frac y x\right)$ with the result adjusted for the correct quadrant.

    Then $\Large \left(\rho e^{\imath \theta}\right)^k = \rho^k e^{\imath k \theta}$

    Also if you have two complex numbers in polar form their product is given by

    $\Large \left(\rho_1 e^{\imath \theta_1}\right)\left(\rho_2 e^{\imath \theta_2}\right)=\rho_1 \rho_2 e^{\imath(\theta_1+\theta_2)}$

    that should be all you need to do your calculation
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: September 24th 2013, 01:16 AM
  2. Replies: 5
    Last Post: January 22nd 2010, 05:50 AM
  3. Replies: 2
    Last Post: August 17th 2008, 12:02 PM
  4. Replies: 3
    Last Post: November 3rd 2007, 01:43 PM
  5. Replies: 0
    Last Post: January 4th 2007, 01:29 PM

Search Tags


/mathhelpforum @mathhelpforum