Results 1 to 4 of 4

Math Help - Complex numbers?? does it go here?

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    Talking Complex numbers?? does it go here?

    solve the equations to find z and w, expressed in a + ib form

    z + (1 - i)w = 2i
    w + (1 - i)z = 1

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by differentiate View Post
    solve the equations to find z and w, expressed in a + ib form

    z + (1 - i)w = 2i
    w + (1 - i)z = 1

    Thanks
    Are z and w meant to represent unknown real numbers? (If so, this is very poor notation since both pronumerals are typically used to represent complex numbers).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    From
    New Delhi
    Posts
    153
    Quote Originally Posted by differentiate View Post
    solve the equations to find z and w, expressed in a + ib form

    z + (1 - i)w = 2i
    w + (1 - i)z = 1

    Thanks
    hello differentiate,
    these are the system of linear equation in two variable w and z(i agree with mr.fantastic that this is wrong representation with complex numbers to use 'z' and 'w' as numeral variables). try to solve them by any method which is suitable to you like substitution method,Elimination method etc.and put i^2=(-1) where require.
    if you have any confusion, look at spoiler;i have used Elimination method in spoiler.

    Spoiler:
    Code:
    
    
    
    
    
     (1 - i)w+z  = 2i \quad ............(1)
    
    
    
    
    
    w + (1 - i)z = 1 \quad ............(2)
    multiply (2) by (1 - i) and subtract from (1), we get
    
    
    
    
    
     \{ (1 - i)w+z\} -\{ (1 - i)w + (1 - i)^2z \}=2i-(1-i)
    
    
    
    
    
     \Rightarrow \quad z-(1 - i)^2z=(3i-1)\quad \Rightarrow \quad z-(-2 i)z=(-1+3i) 
    
    
    
    
    
    \Rightarrow \quad(1+2i)z=-1+3i\quad \Rightarrow \quad z=\frac{-1+3i}{1+2i}
    rationalize 
    
    
    
    
    z=\frac{-1+3i}{1+2i} we get
    
    
    
    
     \boxed{z=1+i}
    substitute value of z in (2),we get 
    
    
    
    
     \boxed{w=-1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    Thanks champions!
    I get it now
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 03:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 25th 2009, 11:22 AM
  4. Replies: 2
    Last Post: February 7th 2009, 06:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum