Results 1 to 10 of 10

Math Help - Solution of a Complex Equation

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    79

    Solution of a Complex Equation

    (3-i)(z+4-2i)=10+20i

    My attempt:

    \begin{array}{rcrcrc}<br />
(3-i)(z+4-2i)=10+20i\\<br />
3z + 12-6i-zi-4i+2i^{2}=10+20i\\<br />
3z+10-10i-zi=10+20i\\<br />
3z-zi=30i<br />
\end{array}

    I have no idea where to go from there.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Viral View Post
    (3-i)(z+4-2i)=10+20i

    My attempt:

    \begin{array}{rcrcrc}<br />
(3-i)(z+4-2i)=10+20i\\<br />
3z + 12-6i-zi-4i+2i^{2}=10+20i\\<br />
3z+10-10i-zi=10+20i\\<br />
3z-zi=30i<br />
\end{array}

    I have no idea where to go from there.
    3z - zi = 30i

    Let z = x + yi \Rightarrow 3z - zi = 3(x+yi) - i(x+yi) = 3x + 3yi - xi + y = (3x + y) + (3y - x)i

    So:

    (3x+y) + (3y-x)i = 0 + 30i

    Can you solve from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    79
    I didn't understand how you worked the following out:

    3x + 3yi - xi + y = (3x + y) + (3y - x)i
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Viral View Post
    I didn't understand how you worked the following out:

    3x + 3yi - xi + y = (3x + y) + (3y - x)i


    3x + 3yi - xi + y = 3x+y+3yi-xi = 3x + y + i(3y) - i(x) = 3x + y + i(3y-x)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    79
    Ehh, this is confusing. I've understood everything so far. So from that I'm guessing 3(x+y)=0, (3y+x)i=30i But I just don't see how I change that into z=x+yi.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Viral View Post
    Ehh, this is confusing. I've understood everything so far. So from that I'm guessing 3(x+y)=0, (3y+x)i=30i But I just don't see how I change that into z=x+yi.
    You are left with 2 equations that you need to solve for x,y:

    3x+y = 0 \Rightarrow x = \frac{-y}{3}
    (3y-x)i = (30)i \Rightarrow 3y - x = 30

    Find the possible values for x,y then substitute them into z = x + yi and get your values of z.

    Spoiler:
    substitute x = -y/3 into the second equation to get:
    3y +y/3 = 30 \Rightarrow \frac{10}{3} \cdot y = 30 \Rightarrow y = 9
    From the first equation we know x = -y/3  \Rightarrow x = -3

    Finally, this leaves us with:
     x = -3, y = 9 \Rightarrow z = x + yi = -3 + 9i
    Last edited by Defunkt; October 10th 2009 at 08:31 AM. Reason: fixed typos
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    79
    How did you get x=-y?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1486
    Awards
    1
    Quote Originally Posted by Viral View Post
    (3-i)(z+4-2i)=10+20i
    You both are making this far too complicated.
    Solve it like an elementary algebra problem: divide and then subtract.

    \frac{{10 + 20i}}{{3 - i}} = \frac{{(10 + 20i)(3 + i)}}{{10}} = 1 + 7i.

    So z=(1+7i)-(4-2i)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2009
    Posts
    79
    \frac{10+20i}{3-i}=(z+4-2i).

    But then how did you get 1+7i from there?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Viral View Post
    \frac{10+20i}{3-i}=(z+4-2i).

    But then how did you get 1+7i from there?
    \frac{10+20i}{3-i} = \frac{10+20i}{3-i} \cdot \frac{3+i}{3+i} = \frac{(10+20i)(3+i)}{(3-i)(3+i)} = \frac{30 + 10i + 60i - 20}{3^2 + 1^2} = \frac{10 + 70i}{10} = 1 + 7i
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Solution Set
    Posted in the Calculus Forum
    Replies: 10
    Last Post: February 6th 2011, 11:15 AM
  2. solution to the equation (complex roots)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 16th 2010, 12:37 PM
  3. [SOLVED] Complex Stiffness Solution
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 5th 2009, 05:16 AM
  4. Complex number solution for theta
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 6th 2008, 08:45 AM
  5. [SOLVED] Complex number solution
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: January 23rd 2008, 09:39 AM

Search Tags


/mathhelpforum @mathhelpforum