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  • Let the complex number z = x + iy be such tha t  is purely imag

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

Option: 1


\frac{3}{2}


Option: 2


\frac{2}{3}


Option: 3

\frac{4}{3}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

z = x + iy

\frac{(2z-3i)}{2z+i}    purely imaginary

Means Re \left(\frac{2 z-3 i}{2 z+i}\right)=0

\begin{aligned} & \Rightarrow \frac{(2 z-3 i)}{(2 z+i)}=\frac{2(x+i y)-3 i}{2(x+i y)+i} \\ & =\frac{2 x+2 y i-3 i}{2 x+i 2 y+i} \\ & =\frac{2 x+i(2 y-3)}{2 x+i(2 y+1)} \times \frac{2 x-i(2 y+1)}{2 x-i(2 y+1)} \end{aligned}

\begin{aligned} & \operatorname{Re}\left[\frac{2 z-3 i}{2 z+i}\right]=\frac{4 x^2+(2 y-3)(2 y+1)}{4 x^2+(2 y+1)^2}=0 \\ & \Rightarrow 4 x^2+(2 y-3)(2 y+1)=0 \\ & \Rightarrow 4 x^2+\left[4 y^2+2 y-6 y-3\right]=0 \\ & \because x+y^2=0 \Rightarrow x=-y^2 \end{aligned}

\begin{aligned} & \Rightarrow 4\left(-y^2\right)^2+4 y^2-4 y-3=0 \\ & \Rightarrow 4 y^4+4 y^2-4 y=3 \\ & \Rightarrow y^4+y^2-y=\frac{3}{4} \end{aligned}

Therefore, correct answer is option (4).

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mansi

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