Results 1 to 6 of 6

Thread: Difficult complex number question

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Difficult complex number question

    The question:
    Let $\displaystyle z = (1 + \sqrt{3} i)$ and w = (1 + i). Find Arg(z) and Arg(w) and hence Arg(zw). Evaluate zw and hence show that $\displaystyle cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$. Find a similar expression for $\displaystyle sin \frac{7\pi}{12}$

    I found:
    Arg(z) = $\displaystyle \frac{\pi}{3}$
    Arg(w) = $\displaystyle \frac{\pi}{4}$
    Arg(zw) = $\displaystyle \frac{\pi}{12}$
    zw = $\displaystyle 2\sqrt{2}e^{i\frac{\pi}{12}}$

    The last bit of the question is giving me trouble. I tried using de Moivre's theorem as follows:

    $\displaystyle (2\sqrt{2}e^{i\frac{\pi}{12}})^7$
    $\displaystyle (2\sqrt{2})^7e^{i\frac{7\pi}{12}}$

    This gives me the argument I want. But I'm unsure of how to solve this. Changing it to sin/cos form (not sure of the proper name for it) seems to be the way to go, but I'm still stumped:

    $\displaystyle (2\sqrt{2})^7(cos(\frac{7\pi}{12}) + isin(\frac{7\pi}{12}))$

    Any assistance would be great!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    I agree with arg(z) and arg(w). However, I don't agree with your arg(zw). How do arguments behave under multiplication?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Oh whoops, that should be 7Pi/12.

    I guess that makes zw:

    $\displaystyle 2\sqrt{2}e^{i\frac{7\pi}{12}}$

    I'm still not sure how to solve it though.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Try multiplying zw in both the polar form and the rectangular form, and then setting them equal. What does that give you?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Ahh! Got it! Thank you very much.

    This is what I did, for anyone that's looking on:

    zw = $\displaystyle (1-\sqrt{3}) + i(1 + \sqrt{3})$ (rectangular form)
    zw = $\displaystyle 2\sqrt{2}(cos(\frac{7\pi}{12}) + isin(\frac{7\pi}{12}))$

    Equating real parts:
    $\displaystyle 2\sqrt{2}(cos(\frac{7\pi}{12})) = 1-\sqrt{3}$
    Therefore,
    $\displaystyle cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Looks good to me. You're very welcome.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Another difficult complex number question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Aug 14th 2010, 02:44 AM
  2. Complex Number Question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 7th 2010, 10:14 PM
  3. complex number question 1
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 9th 2008, 08:02 AM
  4. Complex Number Question.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Apr 6th 2008, 01:07 PM
  5. difficult question on complex analysis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 13th 2006, 04:38 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum