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1. 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?
2. 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
3. (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 and ?? In which case you should know that so they have a real and imaginary part to them, so let and 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.
4. (Original post by RDKGames)
Are you solving for and ?? In which case you should know that so they have a real and imaginary part to them, so let and 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??
5. (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 and

By equating real and imaginary parts:

So you have:

Can you figure out any substitutions to solve these?
6. (Original post by RDKGames)
Let and

By equating real and imaginary parts:

So you have:

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

I said which gives me

also I said which gives me

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.
8. What was the actual question?
9. The complex numbers z and w satisfy the simultaneous equations

2z + iw= -1
z-w =3 + 3i
10. (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.

and divide

.

Updated: April 19, 2017
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