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    2z + iw = -1
    z-w = 3+3i

    I did...

    i(z-w) = i(3+3i)

    zi - wi = 3i -3

    2z + wi + zi - wi = -1 - 3 + 3i

    2z + zi = -4 + 3i

    z = 3

    2z = -4

    this can't be right though, where did I go wrong?
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    You have to remember that z itself is a imaginary number not just a normal number. The answer is really:

    2(a+bi) + i(a+bi) = 3i - 4 where z = a+bi

    This can be rearranged to find z

    2a-b = -4
    a+2b = 3

    If you solve this you get (or at least I did when I tried) z = -1+2i
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    (Original post by staircase)
    2z + iw = -1
    z-w = 3+3i

    I did...

    i(z-w) = i(3+3i)

    zi - wi = 3i -3

    2z + wi + zi - wi = -1 - 3 + 3i

    2z + zi = -4 + 3i

    z = 3

    2z = -4

    this can't be right though, where did I go wrong?
    Are you solving for z and w?? In which case you should know that z,w \in \mathbb{C} so they have a real and imaginary part to them, so let z=a+bi and w=c+di then solve for the 4 variables and you'll have it.

    The reason for why it doesn't work the way you do it is because you assume z and w do not have an imaginary part.
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    (Original post by RDKGames)
    Are you solving for z and w?? In which case you should know that z,w \in \mathbb{C} so they have a real and imaginary part to them, so let z=a+bi and w=c+di then solve for the 4 variables and you'll have it.

    The reason for why it doesn't work the way you do it is because you assume z and w do not have an imaginary part.
    How would you solve it for four variables, I'm unable to get any values through solving for these 4??
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    (Original post by staircase)
    How would you solve it for four variables, I'm unable to get any values through solving for these 4??
    Let z=a+bi and w=c+di

    2z+iw=2(a+bi)+i(c+di)=2a+2bi-d+ci=(2a-d)+(2b+c)i=-1=-1+0i

    By equating real and imaginary parts:

    \Rightarrow 2a-d=-1, 2b+c=0

    z-w=(a+bi)-(c+di)=(a-c)+(b-d)i=3+3i 



\Rightarrow a-c=3, b-d=3

    So you have:

    2a-d=-1

    2b+c=0

    a-c=3

    b-d=3

    Can you figure out any substitutions to solve these?
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    (Original post by RDKGames)
    Let z=a+bi and w=c+di

    2z+iw=2(a+bi)+i(c+di)=2a+2bi-d+ci=(2a-d)+(2b+c)i=-1=-1+0i

    By equating real and imaginary parts:

    \Rightarrow 2a-d=-1, 2b+c=0

    z-w=(a+bi)-(c+di)=(a-c)+(b-d)i=3+3i 



\Rightarrow a-c=3, b-d=3

    So you have:

    2a-d=-1

    2b+c=0

    a-c=3

    b-d=3

    Can you figure out any substitutions to solve these?
    I got a = -11/5, what did you substitute?
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    (Original post by staircase)
    I got a = -11/5, what did you substitute?
    That's wrong.

    I said a=3+c which gives me 2a-d=-1 \Rightarrow 2(3+c)-d=-1 \Rightarrow 2c-d=-7

    also I said b=3+d which gives me 2b+c=0 \Rightarrow 2(3+d)+c=0 \Rightarrow 2d+c=-6

    and now you have 2 equations with the same 2 unknowns that you can solve for, and then you can get a and d from using one of the equations you derived.
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    What was the actual question?
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    The complex numbers z and w satisfy the simultaneous equations

    2z + iw= -1
    z-w =3 + 3i
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    (Original post by staircase)
    2z + iw = -1
    z-w = 3+3i

    I did...

    i(z-w) = i(3+3i)

    zi - wi = 3i -3

    2z + wi + zi - wi = -1 - 3 + 3i

    2z + zi = -4 + 3i
    As an alternative to RDKGames' perfectly correct method you could fix your method..

    You were doing fine up to the point where the quote above ends.

    Now factorise.

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

    and divide

    z=\frac{-4+3i}{2+i}=...=-1+2i.
 
 
 
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