Verified Solution • 2025–26 Session moderate

QJEEM23Macdb8: JEE Main 2023 Mathematics - Class 11 JEEM23Macdb8 Solution: If for , then and are the roots of the equation :

If for z=α+iβ,∣z+2∣=z+4(1+i)z=\alpha+i \beta,|z+2|=z+4(1+i) , then α+β\alpha+\beta and αβ\alpha \beta are the roots of the equation :

Multiple Choice Options

  • A x2+2x−3=0x^{2}+2 x-3=0
  • B x2+3x−4=0x^{2}+3 x-4=0
  • C x2+x−12=0x^{2}+x-12=0
  • D x2+7x+12=0x^{2}+7 x+12=0

Step-by-Step Verified Solution

Given : ∣z+2∣=z+4(1+i)|z+2|=z+4(1+i)

Also, z=α+iβz=\alpha+i \beta

amp;∴∣z+2∣=∣α+iβ+2∣=(α+iβ)+4+4iamp;⇒∣(α+2)+iβ∣=(α+4)+i(β+4)amp;⇒(α+2)2+β2=(α+4)+i(β+4)amp;⇒β+4=0⇒β=−4\begin{aligned}& \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\\\& \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\\\& \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\\\& \Rightarrow \beta+4=0 \Rightarrow \beta=-4\end{aligned}

amp; Now, (α+2)2+β2=(α+4)2amp;⇒α2+4+4α+β2=α2+16+8αamp;⇒4+4α+16=16+8αamp;⇒4α=4⇒α=1amp; So, α+β=−3 and αβ=−4amp;∴ Required equation is amp;x2−(−3−4)x+(−3)(−4)=0amp;⇒x2+7x+12=0\begin{aligned}& \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\\\& \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\\\& \Rightarrow 4+4 \alpha+16=16+8 \alpha \\\\& \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\\\& \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\\\& \therefore \text { Required equation is } \\\\& x^2-(-3-4) x+(-3)(-4)=0 \\\\& \Rightarrow x^2+7 x+12=0\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring