Question

The straight line $(1+2i)z+(2i-1)\bar{z}=10i$ on the complex plane, has intercept on the imaginary axis equal to

Answer 1

Answer: 5

Put $z=iy$, we have
$(1+2i)iy-(2i-1)iy=10i$
$2y+0y=10\Rightarrow y=5$
For $x$-intercept put $z=x+0i\Rightarrow x=5/2$
Alternatively:
Put $z+\bar{z}=0\Rightarrow \bar{z}=-z$
$\Rightarrow (1+2i)z-z(2i-1)=10i$
$\Rightarrow 2z=10i$
$\Rightarrow z=5i;$ or $y=5$