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I have to solve an equation of the type $$a z+b \overline{z}=c$$ with $a,b,c\in\mathbb{C}$.

My approach is to set $F(z)=a z+b \overline{z}-c$ transform $z$ to $x+i y$ and then get a real linear system by assuming $x,y\in\mathbb{R}$ and solve

$$\frac{F(x+i y)+\overline{F(x+i y)}}{2}=0$$ $$\frac{F(x+i y)-\overline{F(x+i y)}}{2i}=0.$$

My problem is that this has to happen a lot for a program that I am writing and I was wondering if there is a faster way to do that. It would be ideal if mathematica can solve these linear systems directly.

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You can use ComplexExpand. –  b.gatessucks Dec 13 '12 at 16:06
I don't really see how this simplifies the procedure. –  tst Dec 13 '12 at 16:11
Was trying to think more generally; will try to prepare an example. –  b.gatessucks Dec 13 '12 at 16:24

2 Answers 2

up vote 8 down vote accepted

Why not just the following?

Solve[a z + b Conjugate[z] == c, z]

Or, if you prefer separating real and complex parts:

Solve[a (x + I y) + b (x - I y) == c, {x, y}]

Or, if you want to worry about degenerate cases, replace Solve above by Reduce in each form.

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Well I have no idea why, I just checked and it works fine. I feel stupid now. –  tst Dec 13 '12 at 16:16
You might need to use some additional conditions, such as specifying that z is an element of Complex... –  tkott Dec 13 '12 at 17:32

As per comment, can use ComplexExpand. Here is one way to go about that.

 ComplexExpand[{Re[a*z + b*Conjugate[z] - c], 
    Im[a*z + b*Conjugate[z] - c]}, {a, b, c, z}] == 0, {Re[z], Im[z]}]

(* {{Re[
    z] -> -((-Im[a] Im[c] + Im[b] Im[c] - Re[a] Re[c] + Re[b] Re[c])/(
    Im[a]^2 - Im[b]^2 + Re[a]^2 - Re[b]^2)), 
  Im[z] -> -((-Im[c] Re[a] - Im[c] Re[b] + Im[a] Re[c] + 
     Im[b] Re[c])/(Im[a]^2 - Im[b]^2 + Re[a]^2 - Re[b]^2))}} *)
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