Complex Number Question 7
Question 7 - 30 January - Shift 1
Let $z=1+i$ and $z_1=\frac{1+i \overline{z}}{\overline{z}(1-z)+\frac{1}{z}}$. Then $\frac{12}{\pi}$ $\arg (z_1)$ is equal to
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Answer: 9
Solution:
Formula: Principal argument, Properties of conjugate, Algebra of complex numbers
$z=1+i$
$z_1=\frac{1+i \overline{z}}{\overline{z}(1-z)+\frac{1}{Z}}$
$z_1=\frac{1+i(1-i)}{(1-i)(1-1-i)+\frac{1}{1+i}}$
$z_1=\frac{1+i-i^{2}}{(1-i)(-i)+\frac{1-i}{2}}$
$z_1=\frac{2+i}{\frac{(-3 i-1)}{2}}=\frac{4+2 i}{-3 i-1}$
$z_1=\frac{-(4+2 i)(3 i-1)}{(3 i)^{2}-(1)^{2}}$
$z_1= -1 + i $
$Arg(z_1)=\frac{3 \pi}{4}$
$\therefore \frac{12}{\pi} \arg (z_1)=\frac{12}{\pi} \times \frac{3 \pi}{4} = 9$
