The argument of the complex number frac{13 - | vedclass

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

Question

(B) Let $z = \frac{13 - 5i}{4 - 9i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4 + 9i)$:$z = \frac{(13

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $z = \frac{13 - 5i}{4 - 9i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4 + 9i)$:$z = \frac{(13 - 5i)(4 + 9i)}{(4 - 9i)(4 + 9i)}$$z = \frac{52 + 117i - 20i - 45i^2}{16 + 81}$$z = \frac{52 + 97i + 45}{97} = \frac{97 + 97i}{97} = 1 + i$The argument of $z = 1 + i$ is $	heta = 	an^{-1}(\frac{1}{1}) = \frac{\pi}{4}$.

The argument of the complex number $\frac{13 - 5i}{4 - 9i}$ is

Similar Questions

If $-\pi < \arg (z) < -\frac{\pi}{2}$, then $\arg (\bar{z}) - \arg (-\bar{z})$ is

Let $z$ and $w$ be complex numbers such that $\overline{z} + i\overline{w} = 0$ and $\text{arg}(zw) = \pi$. Then $\text{arg}(z)$ equals

The amplitude of $\frac{1 + \sqrt{3}i}{\sqrt{3} + i}$ is

The amplitude of $0$ is

The amplitude of the complex number $\frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module