Question

If $z=a+ib$ satisfy $(4+2i)z+(8-2i)z^{\prime }=-2+10i,$ where $z^{\prime }$ is the complex conjugate of $z$, then the value of $|a+2b|$ is

Answer 1

Given : $(4+2i)z+(8-2i)z^{\prime }=-2+10i$
$z=a+ib,z^{\prime }=a-ib$
Now,
$(4+2i)z+(8-2i)z^{\prime }=-2+10i\Rightarrow 4(z+2z^{\prime })+2i(z-z^{\prime })=-2+10i\Rightarrow 4(3a-ib)-4b=-2+10i\Rightarrow (12a-4b)-i\left(4b\right)=-2+10i\Rightarrow 12a-4b=-2,-4b=10\Rightarrow b=-\frac{5}{2},a=-1\therefore |a+2b|=6$