Verified Solution • 2025–26 Session moderate

QJEEM24Ma0206: JEE Main 2024 Mathematics - Class 11 JEEM24Ma0206 Solution: Let be a complex number such that the real part of is zero.…

Let zz be a complex number such that the real part of z−2iz+2i\frac{z-2 i}{z+2 i} is zero. Then, the maximum value of ∣z−(6+8i)∣|z-(6+8 i)| is equal to

Multiple Choice Options

  • A 8
  • B 12
  • C 10
  • D ∞\infty

Step-by-Step Verified Solution

amp;n=z−2iz+2iamp; Let z=x+iyamp;n=x+(y−2)ix+(y+2)i×(x−(y+2)ix−(y+2)i)amp;Re⁡(n)=x2+(y−2)(y+2)x2+(y+2)2=0amp;⇒x2+(y−2)(y+2)=0\begin{aligned}& n=\frac{z-2 i}{z+2 i} \\& \text { Let } z=x+i y \\& n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\& \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\& \Rightarrow x^2+(y-2)(y+2)=0\end{aligned}

amp;⇒x2+y2−4=0amp;⇒x2+y2=4amp; also, ∣z−(6+8i)∣≤∣z∣+∣−6−8i∣amp;∣z−(6+8i)∣≤2+10=12\begin{aligned}& \Rightarrow x^2+y^2-4=0 \\& \Rightarrow x^2+y^2=4 \\& \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\& |z-(6+8 i)| \leq 2+10=12\end{aligned}

Hence, Maximum value of ∣z−(6+8i)∣|z-(6+8 i)| is 12.

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