Let z = 1 + i and z_1 = frac{1 + i overline{z | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $z = 1 + i$,so $\overline{z} = 1 - i$ and $\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$.Substitute these into the expression for $z_1$:$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Given $z = 1 + i$,so $\overline{z} = 1 - i$ and $\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$.Substitute these into the expression for $z_1$:$z_1 = \frac{1 + i(1 - i)}{(1 - i)(1 - (1 + i)) + \frac{1 - i}{2}}$$z_1 = \frac{1 + i - i^2}{(1 - i)(-i) + \frac{1 - i}{2}}$$z_1 = \frac{1 + i + 1}{-i + i^2 + \frac{1 - i}{2}} = \frac{2 + i}{-i - 1 + \frac{1 - i}{2}}$$z_1 = \frac{2 + i}{\frac{-2i - 2 + 1 - i}{2}} = \frac{2(2 + i)}{-1 - 3i} = \frac{2(2 + i)}{-(1 + 3i)}$Multiply numerator and denominator by $(1 - 3i)$:$z_1 = \frac{-2(2 + i)(1 - 3i)}{1^2 + 3^2} = \frac{-2(2 - 6i + i - 3i^2)}{10} = \frac{-2(2 - 5i + 3)}{10} = \frac{-2(5 - 5i)}{10} = -(1 - i) = -1 + i$.Now,find $\arg(z_1)$ for $z_1 = -1 + i$:Since $z_1$ is in the second quadrant,$\arg(z_1) = \pi - 	an^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Finally,calculate $\frac{12}{\pi} \arg(z_1)$:$\frac{12}{\pi} 	imes \frac{3\pi}{4} = 3 	imes 3 = 9$.

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 $..........$.

Similar Questions

Convert the given complex number in polar form: $-1+i$

If $\arg(z) < 0$,then $\arg(-z) - \arg(z)$ is equal to

Argument of the complex number $z = \frac{13-5i}{4-9i}$,where $i = \sqrt{-1}$,is

If $z_1 \cdot z_2 \cdot \dots \cdot z_n = z$,then $arg(z_1) + arg(z_2) + \dots + arg(z_n)$ and $arg(z)$ differ by a

Convert the given complex number in polar form: $i$

Vedclass Test Series

Exam Paper Generator

Online Exam Module