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I can't solve this set of equations, please help me.

$$(1+i)z_1 + (1-i)z_2 = 1+i$$ $$(1-i)z_1 + (1+i)z_2 = 1+3i$$

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Please show what you have done. It's exactly the same process as you would solve a system of equations with real numbers, except that you are computing with complex numbers instead of real numbers. – Ted Oct 23 '11 at 18:08

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One way is to multiply the top equation through by $(1-i)$ and the bottom one by $(1+i)$ to give

$$2z_1 - 2iz_2 = 2\qquad\qquad$$ $$2z_1 + 2iz_2 = -2+4i$$

You can now eliminate one of the unknowns and find the other. You can then substitute this back and get the complete solution.

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Thanks a lot. But I have one more problem. I know answers of this task and I've checked them, but i can't get same answer. Please take a look on this image clip2net.com/clip/m0/1319473344-clip-11kb.png – Ruslan Savenok Oct 24 '11 at 16:23
You have $4z_1=4i$ so $z_1=i$. So $2i-2iz_2=2$, i.e. (dividing by $2i$ on both sides) $1-z_2=-i$, so $z_2=1+i$. – Henry Oct 24 '11 at 18:10
How do you get 4z1 = 4i ? – Ruslan Savenok Oct 24 '11 at 18:39
Added the two equations together. More formally $2iz_2=2z_1-2$ and $2iz_2 = -2 +4i -2z_1$ – Henry Oct 24 '11 at 21:29
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Hints: Gaussian Elimination and Cramer's Rule.

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I'll have complex numbers test and I'm not allowed to use matrix :) – Ruslan Savenok Oct 23 '11 at 18:48
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You don't have to use or mention matrices. Multiply first equation through by $1+i$, the second by $1-i$, subtract. This eliminates $z_2$, and you get a simple linear equation in $z_1$. – André Nicolas Oct 23 '11 at 19:01
What @André suggests is Gaussian Elimination, which does not require matrices. It is one of the ways these problems were solved before matrices. – robjohn♦ Oct 23 '11 at 19:16
@Ruslan Slavenok. It is the way you used to solve two linear equations in two unknowns in early algebra. The only difference is that the numbers are non-real. You can also add the two equations, divide by $2$ to get an equation $z_1+z_2=$ and subtract the second equation from the first, divide by $2i$, to get $z_1-z_2=$. Then I think it will look very familiar. But the first way I described is more general. – André Nicolas Oct 23 '11 at 19:49

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