Results 1 to 5 of 5

Math Help - Complex numbers and polar form

  1. #1
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Complex numbers and polar form

    Hello,

    We're doing a bit of an introduction to complex numbers and their polar form in my math unit. I have a question that according to the unit notes, I have incorrect, but I would think that I am actually, correct. Could someone clarify the correct answer here?

    (2e^{i\frac{\Pi}{5}}})^{25}

    =(2)^{25}(e^{i\frac{\Pi}{5})^{25}

    =2^{25}e^{5i\Pi}

    Now, according to the unit notes, the answer I SHOULD have gotten is

    2^{25}e^{i\Pi}

    Can somebody explain which answer is correct?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2012
    From
    Lima, Perú
    Posts
    11
    Thanks
    1

    Re: Complex numbers and polar form

    your answer is correct too...

    Remember that:

    \mathrm{Z=e^{i\theta}=\cos(\theta)+i\sin(\theta)=\  cos(2k\pi+\theta)+i\sin(2k\pi+\theta)=e^{i(2k\pi+\  theta)}\qquad k \in \mathds{Z}}
    Last edited by darthjavier; November 18th 2012 at 06:27 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294

    Re: Complex numbers and polar form

    Your answer is INCORRECT, because we define complex numbers in the region \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}. If you end up with an angle outside that region you need to add/subtract multiples of \displaystyle \begin{align*} 2\pi \end{align*} until you ARE in that region.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2012
    From
    Lima, Perú
    Posts
    11
    Thanks
    1

    Re: Complex numbers and polar form

    Quote Originally Posted by Prove It View Post
    Your answer is INCORRECT, because we define complex numbers in the region \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}. If you end up with an angle outside that region you need to add/subtract multiples of \displaystyle \begin{align*} 2\pi \end{align*} until you ARE in that region.
    but the two answers are okay, right?

    As far as I've understood it is a convention to not have two identical answers. I'm not sure...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294

    Re: Complex numbers and polar form

    Quote Originally Posted by darthjavier View Post
    but the two answers are okay, right?

    As far as I've understood it is a convention to not have two identical answers. I'm not sure...
    The two answers are identical, but you are EXPECTED to use the convention that \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*} to make life easier for everyone.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: July 27th 2012, 06:40 PM
  2. Complex Numbers in Polar Form
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 19th 2011, 09:36 AM
  3. more complex numbers help - polar form.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 16th 2011, 12:06 PM
  4. Polar form of complex numbers
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: April 9th 2010, 06:26 AM
  5. complex numbers in polar form
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 8th 2008, 04:00 AM

Search Tags


/mathhelpforum @mathhelpforum