Verified Solution • 2025–26 Session moderate QJEEM23Maaf18: JEE Main 2023 Mathematics - Class 11 JEEM23Maaf18 Solution: Let and . Then is equal to __________. Let z=1+iz=1+iz=1+i and z1=1+izˉzˉ(1−z)+1zz_{1}=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}z1=zˉ(1−z)+z11+izˉ . Then 12πarg(z1)\frac{12}{\pi} \arg \left(z_{1}\right)π12arg(z1) is equal to __________. Step-by-Step Verified Solution z=1+iz = 1 + iz=1+i z1=1+iz‾z‾(1−z)+1z{z_1} = {{1 + i\overline z } \over {\overline z (1 - z) + {1 \over z}}}z1=z(1−z)+z11+iz =z(1+iz‾)∣z∣2(1−z)+1= {{z(1 + i\overline z )} \over {|z{|^2}(1 - z) + 1}}=∣z∣2(1−z)+1z(1+iz) =(1+i)(1+i(1−i))2(1−1−i)+1= {{(1 + i)(1 + i(1 - i))} \over {2(1 - 1 - i) + 1}}=2(1−1−i)+1(1+i)(1+i(1−i)) z1=1−i{z_1} = 1 - iz1=1−i argz1=tan−1(−11)=3π4\arg {z_1} = {\tan ^{ - 1}}\left( {{{ - 1} \over 1}} \right) = {{3\pi } \over 4}argz1=tan−1(1−1)=43π 12πarg(z1)=3π4 . 12π=9{{12} \over \pi }\arg ({z_1}) = {{3\pi } \over 4}\,.\,{{12} \over \pi } = 9π12arg(z1)=43π.π12=9 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