JEE Main 20248 Apr, Morning shiftMathematicsMediumComplex Numbers and Quadratic Equations264× since 2002Let zzz be a complex number such that ∣z+2∣=1|z+2|=1∣z+2∣=1 and lm(z+1z+2)=15\operatorname{lm}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}lm(z+2z+1)=51. Then the value of ∣Re(z+2‾)∣|\operatorname{Re}(\overline{z+2})|∣Re(z+2)∣ isAOption A. 265\frac{2 \sqrt{6}}{5}526BOption B. 245\frac{24}{5}524COption C. 65\frac{\sqrt{6}}{5}56DOption D. 1+65\frac{1+\sqrt{6}}{5}51+6CheckSkip · show solution
