If {z_1} and {z_2} are two non-zero complex n | vedclass

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

Question

(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$. This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originat

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$. This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originating from the origin in the complex plane. Let ${z_1} = {r_1}(\cos{	heta_1} + i\sin{	heta_1})$ and ${z_2} = {r_2}(\cos{	heta_2} + i\sin{	heta_2})$. Then $|{z_1} + {z_2}|^2 = (|{z_1}| + |{z_2}|)^2 = |{z_1}|^2 + |{z_2}|^2 + 2|{z_1}||{z_2}|$. Also,$|{z_1} + {z_2}|^2 = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2 = |z_1|^2 + |z_2|^2 + 2	ext{Re}(z_1\overline{z_2})$. Comparing these,$2	ext{Re}(z_1\overline{z_2}) = 2|z_1||z_2|$,which means $\cos(	heta_1 - 	heta_2) = 1$. Thus,$	heta_1 - 	heta_2 = 0$,which implies $	ext{arg}({z_1}) - 	ext{arg}({z_2}) = 0$.

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$ then $\text{arg}({z_1}) - \text{arg}({z_2})$ is equal to

Similar Questions

If $Arg(z)$ denotes the principal argument of a complex number $z$,then the value of the expression $Arg\left( -i e^{i\frac{\pi}{9}} z^2 \right) + 2Arg\left( 2i e^{-i\frac{\pi}{18}} \bar{z} \right)$ is

Let the two values of $z = \sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2} < \operatorname{Arg}(z_1) < \operatorname{Arg}(z_2) < \pi$,then $\arg(z_1) + \arg(z_2) = $

If $z_1=(2,-1)$ and $z_2=(6,3)$,then $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right)=$

If $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$,then $arg(z) = $ ............. $^\circ$

The amplitude of $\sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module