help with system with 2 complex unknowns?

Question

i have this system where z,w belong to C. Z=x+yi, W=a+bi
Because i cannot use a symbol for the x-yi or a-bi i will use the symbol z* and w*. So there it is:
(1+2i)z+(3-i)w=4i
(3+2i)z*-5iw*=-1-11i

How do i solve it easily??

My mistake Z=z. and W=w

and i say that if z= x+yi and if w=a+bi
then z*=x-yi then w*=a-bi

O.K done with Mathsmanretired way, but i was just wondering if there is another way of solving. Nevermind , i can solve it so there is no problem thank you guys...

Answer 1

Someone may come up with a faster way than this but . . .
If you replace z, w, z*, w* by x + yi, x - yi, a + bi, a - bi
and multiply out all the brackets you will have two complex equations. Each one can be split into two by equating real parts and imaginary parts. This gives you four equations in four real unknowns which you should be able to solve.
This won't be easy so I hope that you are given a faster way.
Will come back if I think of one.

Edit. Actually this way is not too difficult. If you do it correctly
you will find that a, b, x, y are 0, 1, 1, 2 but not necessarily in that order!

Answer 2

if z <> z* and w <> w*, you have 4 variables, not 2. if they ARE equal, why confuse things with extra symbols? it looks like z* is the conjugate of Z, but is Z = z? the ambiguities are probably the reason no one's answering the question.

Answer 3

To simplify, maybe you can use:

z + z* = 2x
z - z* = 2yi
w + w* = 2a
w - w* = 2bi