Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Hi, having problems with a complex number multiplication

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    uk
    Posts
    5

    Hi, having problems with a complex number multiplication

    Hi everyone, I'm Joanne and currently studying to become a maths teacher (1 year in to studying).


    I'm having problems trying to work out the below multiplication.

    Hi, having problems with a complex number multiplication-maths.gif


    I've worked out that the polar form of (2-2i)^12 is Hi, having problems with a complex number multiplication-math_image-7-.gif ( sorry about the ?'s)

    But I am unsure of what to do next.

    Do I work out the polar form of Hi, having problems with a complex number multiplication-math_image-8-.gif, then multiply them together, or am I doing this wrong to begin with?


    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: Hi, having problems with a complex number multiplication

    Quote Originally Posted by Jojo55 View Post
    Hi everyone, I'm Joanne and currently studying to become a maths teacher (1 year in to studying).
    I'm having problems trying to work out the below multiplication.
    Click image for larger version. 

Name:	maths.gif 
Views:	22 
Size:	671 Bytes 
ID:	27644
    I've worked out that the polar form of (2-2i)^12 is Click image for larger version. 

Name:	math_image (7).gif 
Views:	1 
Size:	1.9 KB 
ID:	27645 ( sorry about the ?'s)

    (2-2i)=2\sqrt 2 \exp \left( {\frac{{ -i \pi }}{4}} \right).

    thus (2-2i)^{12}=\left(2\sqrt 2\right)^{12} \exp \left( -3i\pi \right).

    But \left(2\sqrt 2\right)^{12}=2^{18}.

    Can you finish?
    Last edited by Plato; March 22nd 2013 at 11:05 AM.
    Thanks from Jojo55
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2013
    From
    Iowa, USA
    Posts
    37
    Thanks
    9

    Re: Hi, having problems with a complex number multiplication

    Are you familiar with the Binomial Theorem?

    Study this link: Binomial Theorem
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: Hi, having problems with a complex number multiplication

    Quote Originally Posted by jpritch422 View Post
    Are you familiar with the Binomial Theorem?

    Study this link: Binomial Theorem

    What does that have to do with the OP?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Henderson's Avatar
    Joined
    Dec 2007
    Posts
    127
    Thanks
    2

    Re: Hi, having problems with a complex number multiplication

    Quote Originally Posted by Plato View Post
    What does that have to do with the OP?

    I think jpritch is suggesting that the problem be done without DeMoivre/Polar form.
    I disagree with that suggestion, but I'm fairly certain that's where he's going with that.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: Hi, having problems with a complex number multiplication

    Quote Originally Posted by Henderson View Post
    I think jpritch is suggesting that the problem be done without DeMoivre/Polar form.
    I disagree with that suggestion, but I'm fairly certain that's where he's going with that.
    Actually, I understood what the post meant. I want to see why jpritch422 wanted to use that.

    The irony is that DeMoivre's Rule is used for many proofs about binomial coefficients.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2013
    From
    uk
    Posts
    5

    Re: Hi, having problems with a complex number multiplication

    Quote Originally Posted by Plato View Post
    (2-2i)=2\sqrt 2 \exp \left( {\frac{{ -i \pi }}{4}} \right).

    thus (2-2i)^{12}=\left(2\sqrt 2\right)^{12} \exp \left( -3i\pi \right).

    But \left(2\sqrt 2\right)^{12}=2^{18}.

    Can you finish?
    Sorry for the late reply. I managed to solve to get the answer of -1

    Thanks for the help on this
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 25th 2012, 03:14 PM
  2. Two complex number problems!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 30th 2010, 06:36 AM
  3. A few complex number problems
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 1st 2010, 08:02 AM
  4. Complex number plane multiplication
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 28th 2010, 06:02 PM
  5. Two complex number problems!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 2nd 2009, 11:49 AM

Search Tags


/mathhelpforum @mathhelpforum