If $(3+i)z=(3-i)\overline{z},$ then complex number z is

Given :$(3+i)z=(3-i)\overline{z}$

Let $z=x(3-i), x\in R$

L.H.S. $=(3+i)z =(3+i)x(3-i)$

$=x(3+i)(3-i)=x\left[{\left(3\right)}^2+1^2\right]=10x$

R.H.S. = $(3-i)\overline{z}=(3-i)x(3+i)=x[3^2+1^2]=10x$

Hence, L.H.S. = R.H.S.

$\because z=x(3-i)$ satisfies the equation, then

$z=x(3-i),$ where x is a real number.