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How to solve the equation like this:

$-par-(x+yi)-a\overline{r}=0 $ where $p,a,x,y$ are constant real numbers, $r$ is complex number and $\overline{r}$ is its complex conjugate. I want to solve for $r$

I tried to use Assumptions on $p,a,x,y$ so that they are real and tried to solve it, but I get a very weird solution, i.e

Assumptions = {{a,p} [Element] Reals}
Solve[-r a p - (x+yI) - a r[Conjugate] == 0 ,r]

which is different from the answer in the book. Any guidance for this?

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Is this what you are looking for?

ClearAll[r, a, p, x, y];
ComplexExpand@Solve[-r a p - (x + y*I) - a *Conjugate[r] == 0, r]

enter image description here

which is different from the answer in the book.

What is the book answer?

Also in Mathematica be careful with space. yI is not the same as y*I.

Also You can use ComplexExpand instead of assuming variables are real.

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  • $\begingroup$ Yes! this is the answer. What is meant by ClearAll anyway? $\endgroup$
    – Nothing
    Jun 12 at 11:53
  • $\begingroup$ @Nothing it is always good idea to clear variables before using them. $\endgroup$
    – Nasser
    Jun 12 at 11:54

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