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

Complex Numbers and Quadratic Equations

Solution:

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