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4-1.Complex numbers
hard
माना $\mathrm{z}=1+\mathrm{i}$ तथा $\mathrm{z}_1=\frac{1+\mathrm{i} \overline{\mathrm{z}}}{\overline{\mathrm{z}}(1-\mathrm{z})+\frac{1}{\mathrm{z}}}$ है तो $\frac{12}{\pi} \arg \left(\mathrm{z}_1\right)$ बराबर है____________.
A
$18$
B
$27$
C
$36$
D
$9$
(JEE MAIN-2023)
Solution
$z=1+i$
$z_1=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$
$z_1=\frac{1+i(1-i)}{(1-i)(1-1-i)+\frac{1}{1+i}}$
$=\frac{1+i-i^2}{(1-i)(-i)+\frac{1-i}{2}}$
$=\frac{2+i}{-3 i-1}=\frac{4+2 i}{-3 i-1}$
$=\frac{-(4+2 i)(3 i-1)}{(3 i)^2-(1)^2}$
$\therefore \frac{12}{\pi} \arg \left(z_1\right)=\frac{12}{\pi} \times \frac{3 \pi}{4}=9$
Standard 11
Mathematics
