Results 1 to 10 of 10

Thread: Solution of a Complex Equation

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    79

    Solution of a Complex Equation

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

    My attempt:

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

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

    My attempt:

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

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

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

    So:

    $\displaystyle (3x+y) + (3y-x)i = 0 + 30i$

    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:

    $\displaystyle 3x + 3yi - xi + y = (3x + y) + (3y - x)i$
    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:

    $\displaystyle 3x + 3yi - xi + y = (3x + y) + (3y - x)i$


    $\displaystyle 3x + 3yi - xi + y = 3x+y+3yi-xi = 3x + y + i(3y) - i(x) = 3x + y + i(3y-x)$
    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 $\displaystyle 3(x+y)=0, (3y+x)i=30i$ But I just don't see how I change that into $\displaystyle z=x+yi$.
    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 $\displaystyle 3(x+y)=0, (3y+x)i=30i$ But I just don't see how I change that into $\displaystyle z=x+yi$.
    You are left with 2 equations that you need to solve for x,y:

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

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

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

    Finally, this leaves us with:
    $\displaystyle x = -3, y = 9 \Rightarrow z = x + yi = -3 + 9i$
    Last edited by Defunkt; Oct 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 $\displaystyle x=-y$?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Quote Originally Posted by Viral View Post
    $\displaystyle (3-i)(z+4-2i)=10+20i$
    You both are making this far too complicated.
    Solve it like an elementary algebra problem: divide and then subtract.

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

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

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

    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
    $\displaystyle \frac{10+20i}{3-i}=(z+4-2i)$.

    But then how did you get 1+7i from there?
    $\displaystyle \frac{10+20i}{3-i} = \frac{10+20i}{3-i} \cdot \frac{3+i}{3+i} = $ $\displaystyle \frac{(10+20i)(3+i)}{(3-i)(3+i)} = \frac{30 + 10i + 60i - 20}{3^2 + 1^2} = $ $\displaystyle \frac{10 + 70i}{10} = 1 + 7i$
    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: Feb 6th 2011, 11:15 AM
  2. solution to the equation (complex roots)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jan 16th 2010, 12:37 PM
  3. [SOLVED] Complex Stiffness Solution
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Nov 5th 2009, 05:16 AM
  4. Complex number solution for theta
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jun 6th 2008, 08:45 AM
  5. [SOLVED] Complex number solution
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Jan 23rd 2008, 09:39 AM

Search Tags


/mathhelpforum @mathhelpforum