Let z be a complex number such that |z+2|=1 and Im((z+1)/(z+2))=(1)/(5) . Then the value of |Re( z+2 )| is

Correct answer: A. 2 6 5 |z+2|=1, Im((z+1)/(z+2))=(1)/(5) Let z+2= θ+i θ aligned & (1)/(z+2)= θ-i θ & (z+1)/(z+2)=1-(1)/(z+2)=1-( θ-i θ) & =(1- θ)+isin θ & Im((z+1)/(z+2))= θ, θ=(1)/(5) & θ= ± 1-(1)/(25) = ± 2 6 5 & |Re( z+2 )|= 2 6 5 aligned

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Complex Number

Let z be a complex number such that |z+2|=1 and...

Let

z

be a complex number such that

|z+2|=1

and

\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}

. Then the value of

|\operatorname{Re}(\overline{z+2})|

Options

A $\frac{2 \sqrt{6}}{5}$
B $\frac{24}{5}$
C $\frac{1+\sqrt{6}}{5}$
D $\frac{\sqrt{6}}{5}$

Correct answer

A. $\frac{2 \sqrt{6}}{5}$

Step-by-step solution

|z+2|=1, \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}

Let

z+2=\cos \theta+i \sin \theta

\begin{aligned} & \frac{1}{z+2}=\cos \theta-i \sin \theta \\ & \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) \\ & =(1-\cos \theta)+\operatorname{isin} \theta \\ & \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} \\ & \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \\ & |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5} \end{aligned}

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