Results 1 to 3 of 3

Math Help - Complex numbers: equation

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    3

    Complex numbers: equation

    Could enyone solve this for me:

    Equation system:

    (−2 + 2i)z1 + (3 + 3i)z2 = 6 + 2i
    (3 + 4i)z1 + (8 − 6i)z2 = −6 + 17i

    Im(z1)
    Re(z1)
    Im(z2)
    Re(z2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2007
    Posts
    21
    Quote Originally Posted by clapaucius View Post
    Could enyone solve this for me:

    Equation system:

    (-2 + 2i)z1 + (3 + 3i)z2 = 6 + 2i
    (3 + 4i)z1 + (8 - 6i)z2 = -6 + 17i

    Im(z1)
    Re(z1)
    Im(z2)
    Re(z2)

    (-2 + 2i)z1 + (3 + 3i)z2 = 6 + 2i
    --> z1 = [(6+2i)-(3+3i)z2]/(-2+2i)

    (3 + 4i)z1 + (8 - 6i)z2 = -6 + 17i
    (3+4i)[(6+2i)-(3+3i)z2]/(-2+2i) + (8-6i)z2 = (-6+17i)
    [ (3+4i)(6+2i)-(3+4i)(3+3i)z2 ] / (-2+2i) + (8-6i)z2 = (-6+17i)
    [ (10+30i)-(-3+21i)z2 ] / (-2+2i) + (8-6i)z2 = (-6+17i)
    (10+30i)/(-2+2i) - (-3+21i)z2/(-2+2i) + [(-2+2i)(8-6i)z2]/(-2+2i) = (-6+17i)
    z2 = [ (-6+17i)-[(10+30i)/(-2+2i)] ] * (-2+2i)/[(-3+21i)+(-2+2i)(8-6i)
    z2 = [ (-6+17i)(-2+2i)-(10+30i) ]/(-2+2i) ] * (-2+2i)/(-7+49i)
    z2 = [ (-6+17i)(-2+2i)-(10+30i) ]/(-7+49i)
    z2 = [ (-22-46i)-(10+30i) ]/(-7+49i)
    z2 = (-12-76i)/(-7+49i)
    z2 = (12+76i)/(7-49i)
    z2 = (12+76i)/(7-49i) * (7+49i)/(7+49i)
    z2 = (-3640+1120i)/(-2352)
    z2 = (-3640+1120i)/(-2352)
    z2 = (-3640/-2352) + i(1120/2352)

    Therefore Re(z2)=(3640/2352) and Ir(z2)=(1120/2352)
    (You can simplify fractions).

    To find Re(z1) and Ir(z1) plug z2 = (3640/2352) + i(1120/2352) back into either of the two original equations and solve.

    !

    Note: I would check all my work, since the probability of a mistake is extremely high. Hand solving is much easier.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,646
    Thanks
    1596
    Awards
    1
    \begin{array}{l}<br />
 \left( { - 2 + 2i} \right)z_1  + \left( {3 + 3i} \right)z_2  = 6 + 2i \\ <br />
 \left( {3 + 4i} \right)z_1  + \left( {8 - 6i} \right)z_2  =  - 6 + 17i \\ <br />
 \end{array}

    Here are some useful equations. <br />
\begin{array}{l}<br />
 \Re e(z + w) = \Re e(z) + \Re e(w) \\ <br />
 \Im m(z + w) = \Im m(z) + \Im m(w) \\ <br />
 \Re e(zw) = \Re e(z)\Re e(w) - \Im m(z)\Im m(w) \\ <br />
 \Im m(zw) = \Re e(z)\Im m(w) + \Re e(w)\Im m(z) \\ <br />
 \end{array}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 23rd 2011, 09:49 PM
  2. [SOLVED] Complex Numbers [Equation]
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 12th 2011, 11:51 AM
  3. Equation with complex numbers
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: December 9th 2010, 07:07 AM
  4. complex numbers equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 18th 2009, 01:10 AM
  5. complex numbers equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 14th 2009, 02:54 PM

Search Tags


/mathhelpforum @mathhelpforum