Results 1 to 10 of 10

Math Help - complex calculation

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    12

    complex calculation

    Hi, can some one help me please solve that?
    Let  v=1-\sqrt{3}i , u=i-1
    compute v^6u^5

    So i did:
    \phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi
    r=\sqrt{1^2+\sqrt{3}^2} =2
    \Rightarrow v=2e^{\frac{\pi}{3}i}

    \phi_2 = tan^{-1}(\frac{1}{-1})=-\frac{1}{4}\pi = \frac{3}{4}\pi
    r_2=\sqrt{1^2+1^2} =\sqrt{2}
    \Rightarrow u=\sqrt{2}e^{\frac{3\pi}{4}i}
    now i did:
    (2e^{\frac{\pi}{3}i})^6 (\sqrt{2}e^{\frac{3\pi}{4}i})^5

    2^62^{\frac{5}{2}}(e^{\frac{\pi}{3}i})^6(e^{\frac{  3\pi}{4}i})^5

    2^62^{\frac{5}{2}}(e^{\frac{6\pi}{3}i})(e^{\frac{1  5\pi}{4}i})

    2^{6+\frac{5}{2}}(e^{\frac{6\pi}{3}i+\frac{15\pi}{  4}i})

    2^{\frac{17}{2}}e^{\frac{23\pi}{4}i}

    am i right?
    I think I have a mistake somewhere...
    If not, how can i go from here?

    Thanks!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    23Pi/4 can be simplified to 7Pi/4 or -Pi/4

    Where do you want to go with it?

    \displaystyle re^{i\theta}=r(\cos{\theta}+i\sin{\theta})
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    12
    to r=a+bi
    i did not did mistakes?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    It is correct. Is there a certain form you want to represent it in? You can simplify the angle (post #2) but that is about it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Nati View Post
    to r=a+bi
    i did not did mistakes?
    r=\sqrt{a^2+b^2}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    You mean in the form z = a+bi

    I get \displaystyle (1-\sqrt{3}i)^6\times (-1+i)^5 = 64\times (4-4i)
    Last edited by pickslides; January 2nd 2011 at 08:38 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    12
    Quote Originally Posted by pickslides View Post
    You mean in the form z = a+bi

    I get \displaystyle (1-\sqrt{3})^6\times (-1+i)^5 = 64\times (4-4i)
    Can you share your way please?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    I use De Moivre's theorem

    \displaystyle (x+yi)^n= r^n(\cos n\theta +i \sin n\theta)

    Give it a go!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Nati View Post
    Hi, can some one help me please solve that?
    Let  v=1-\sqrt{3}i , u=i-1
    compute v^6u^5

    So i did:
    \phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi
    r=\sqrt{1^2+\sqrt{3}^2} =2
    \Rightarrow v=2e^{\frac{\pi}{3}i}
    The above should be: \displaystyle v=2e^{-\frac{\pi}{3}i}.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Nati View Post
    to r=a+bi
    i did not did mistakes?
    You were told that your calculation ws correct but that the agument could be simplified. Now it looks like you want to give your answer in cartesian form. You should know how to convert from polar to cartesian form. Where are you stuck?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: December 16th 2011, 03:19 AM
  2. Complex Commission Calculation Question
    Posted in the Business Math Forum
    Replies: 4
    Last Post: August 2nd 2011, 08:47 PM
  3. [SOLVED] Question about Complex Number Calculation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 28th 2011, 08:18 PM
  4. GDP calculation
    Posted in the Business Math Forum
    Replies: 0
    Last Post: December 1st 2010, 10:33 AM
  5. SST Calculation
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: October 21st 2010, 01:58 PM

Search Tags


/mathhelpforum @mathhelpforum