Let the complex number z=x+iy be such that (2z-3i)/(2z+i) is purely imaginary.

Question

Let the complex number $z=x+iy$ be such that $\frac{2z-3i}{2z+i}$ is purely imaginary. If $x+{y}^{2}=0$, then ${y}^{4}+{y}^{2}-y$ is equal to :

Accepted Answer

Let, $z=x+iy$

So, $\frac{2z-3i}{2z+i}$ is purely imaginary
[Given]

$\begin{gathered} (\frac{2z-3i}{2z+i})+(\frac{\vec{2z-3i}}{2z+i})=0 \ \Rightarrow \frac{2z-3i}{2z+i}+\frac{2\vec{z}+3i}{2\vec{z}-i}=0 \end{gathered}$

$\begin{gathered} \Rightarrow (2z-3i)(2\vec{z}-i)+(2\vec{z}+3i)(2z+i)=0 \ \Rightarrow {4|z{|}^{2}-6i\vec{z}-2iz-3}+{4|z{|}^{2}+6iz+2i\vec{z}-3}=0 \ \Rightarrow 8|z{|}^{2}-4i\vec{z}+4iz-6=0 \ \Rightarrow 8({x}^{2}+{y}^{2})-4i(x-iy)+4i(x+iy)-6=0 \ \Rightarrow 8({x}^{2}+{y}^{2})-4ix-4y+4ix-4y-6=0 \ \Rightarrow 8({x}^{2}+{y}^{2})-8y-6=0 \ \Rightarrow 4({x}^{2}+{y}^{2})-4y-3=0 \end{gathered}$

Given that, $x+{y}^{2}=0$

$\begin{gathered} \Rightarrow x=-{y}^{2} \ \Rightarrow 4({y}^{4}+{y}^{2})-4y=3 \ \Rightarrow {y}^{4}+{y}^{2}-y=\frac{3}{4} \end{gathered}$

Hence, required answer is $\frac{3}{4}$.

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