
Complex number question
Find Z1 and Z2 , given that:
(1 + i )Z1 + Z2 = 1 + i
4Z1 + (1  i)Z2 = 4+ 2i
I tried to solve by letting Z1 = a + bi and Z2 = c + di and comparing coefficients with the Cartesian form on the right hand side of the equation. I cant get the answer as solving four unknowns seems a bit too much for me, is there any simpler way to solve this??

$\displaystyle \left\{\begin{array}{l}(1+i)z_1+z_2=1+i\\4z_1+(1i)z_2=4+2i\end{array}\right.$
Multiply the first equation with 4 and the second with (1+i). Then add the equations:
$\displaystyle 2z_2=22i\Rightarrow z_2=1i$
Plug $\displaystyle z_2$ in the second equation:
$\displaystyle 4z_1=4+4i\Rightarrow z_1=1+i$

Thanks! I got an idea similar as yours earlier but still cant get it to work, now i know why. what i was think about just now is letting z1 = ( 1  i) and letting z2 = (1+i), obviously that is the wrong substitution. Thanks again!

On the other hand, if you multiply the first equation by $\displaystyle 1i$ we get $\displaystyle 2z_1+(1i)z_2=2$.
Now substract that from the second equation.