JEE Main 2023 — Mathematics Algebra

Given:

$$z=1+i$$z=1+i

$$\bar{z}=1-i$$zˉ=1−i

And, $\frac{1}{z}=\frac{1}{1+i}\times \frac{1-i}{1-i}=\frac{1-i}{2}$z1​=1+i1​×1−i1−i​=21−i​

So,

$${z}_{1}=\frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$$z1​=zˉ(1−z)+z1​1+izˉ​

$$=\frac{1+i-{i}^{2}}{(1-i)(-i)+\frac{1-i}{2}}$$=(1−i)(−i)+21−i​1+i−i2​

$$=\frac{1+i-{i}^{2}}{-i+{i}^{2}+\frac{1}{2}-\frac{i}{2}}$$=−i+i2+21​−2i​1+i−i2​

$$=\frac{2+i}{\frac{-3i-1}{2}}=\frac{4+2i}{-3i-1}$$=2−3i−1​2+i​=−3i−14+2i​

$$=\frac{-(4+2i)(3i-1)}{(3i{)}^{2}-(1{)}^{2}}$$=(3i)2−(1)2−(4+2i)(3i−1)​

$$=\frac{12i-4+6{i}^{2}-2i}{10}$$=1012i−4+6i2−2i​

$$=\frac{10i-10}{10}=i-1$$=1010i−10​=i−1

$$Arg({z}_{1})=\frac{3\pi }{4}$$Arg(z1​)=43π​

$$\therefore \frac{12}{\pi }\mathrm{arg}({z}_{1})=\frac{12}{\pi }\times \frac{3\pi }{4}=9$$∴π12​arg(z1​)=π12​×43π​=9