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QJEEM23Ma9b78: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9b78 Solution: Let the complex number be such that is purely imaginary. If ,…

Let the complex number z=x+iyz = x + iy be such that 2z−3i2z+i{{2z - 3i} \over {2z + i}} is purely imaginary. If x+y2=0{x} + {y^2} = 0 , then y4+y2−y{y^4} + {y^2} - y is equal to :

Multiple Choice Options

  • A 43{4 \over 3}
  • B 32{3 \over 2}
  • C 34{3 \over 4}
  • D 23{2 \over 3}

Step-by-Step Verified Solution

Let, z=x+iyz=x+i y

So, 2z−3i2z+i\frac{2 z-3 i}{2 z+i} is purely imaginary   ~~ [Given]

amp; Now, (2z−3i2z+i)+(2z−3i‾2z+i)=0amp;⇒2z−3i2z+i+2zˉ+3i2zˉ−i=0\begin{aligned}& \text { Now, }\left(\frac{2 z-3 i}{2 z+i}\right)+\left(\frac{\overline{2 z-3 i}}{2 z+i}\right)=0 \\\\& \Rightarrow \frac{2 z-3 i}{2 z+i}+\frac{2 \bar{z}+3 i}{2 \bar{z}-i}=0\end{aligned}

amp;⇒(2z−3i)(2zˉ−i)+(2zˉ+3i)(2z+i)=0amp;⇒{4∣z∣2−6izˉ−2iz−3}+{4∣z∣2+6iz+2izˉ−3}=0amp;⇒8∣z∣2−4izˉ+4iz−6=0amp;⇒8(x2+y2)−4i(x−iy)+4i(x+iy)−6=0amp;⇒8(x2+y2)−4ix−4y+4ix−4y−6=0amp;⇒8(x2+y2)−8y−6=0amp;⇒4(x2+y2)−4y−3=0\begin{aligned}& \Rightarrow(2 z-3 i)(2 \bar{z}-i)+(2 \bar{z}+3 i)(2 z+i)=0 \\\\& \Rightarrow\left\{4|z|^2-6 i \bar{z}-2 i z-3\right\}+\left\{4|z|^2+6 i z+2 i \bar{z}-3\right\}=0 \\\\& \Rightarrow 8|z|^2-4 i \bar{z}+4 i z-6=0 \\\\& \Rightarrow 8\left(x^2+y^2\right)-4 i(x-i y)+4 i(x+i y)-6=0 \\\\& \Rightarrow 8\left(x^2+y^2\right)-4 i x-4 y+4 i x-4 y-6=0 \\\\& \Rightarrow 8\left(x^2+y^2\right)-8 y-6=0 \\\\& \Rightarrow 4\left(x^2+y^2\right)-4 y-3=0\end{aligned}

Given that, x+y2=0x+y^2=0

amp;⇒x=−y2amp;⇒4(y4+y2)−4y=3amp;⇒y4+y2−y=34\begin{aligned}& \Rightarrow x=-y^2 \\\\& \Rightarrow 4\left(y^4+y^2\right)-4 y=3 \\\\& \Rightarrow y^4+y^2-y=\frac{3}{4}\end{aligned}

Hence, required answer is 34\frac{3}{4} .

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