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

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

Question

(d) Let ${z_1} = {r_1}$$(\cos {	heta _1} + i\sin {	heta _1})$,${z_2} = {r_2}$ $(\cos {	heta _2} + i\sin {	heta _2})$
$	herefore $$|{z_1} + {z_2}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(d) Let ${z_1} = {r_1}$$(\cos {	heta _1} + i\sin {	heta _1})$,${z_2} = {r_2}$ $(\cos {	heta _2} + i\sin {	heta _2})$
$	herefore $$|{z_1} + {z_2}| = [{({r_1}\cos {	heta _1} + {r_2}\cos {	heta _2})^2}$
$ + {({r_2}\sin {	heta _1} + {r_2}\sin {	heta _2})^2}{]^{1/2}}$
$ = {[r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({	heta _1} - {	heta _2})]^{1/2}} = {[{({r_1} + {r_2})^2}]^{1/2}}$

$|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$
Therefore $\cos ({	heta _1} - {	heta _2}) = 1 \Rightarrow {	heta _1} - {	heta _2} = 0 \Rightarrow {	heta _1} = {	heta _2}$
Thus arg $({z_1}) - arg({z_2}) = 0$.
Trick : $|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}| \Rightarrow {z_1},{z_2}$lies on same straight line.
$	herefore arg\,{z_1} = arg\,{z_2} \Rightarrow arg\,{z_1} - 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 arg $({z_1}) - $arg $({z_2})$ is equal to

Similar Questions

If $z = 3 + 5i,\,\,{\rm{then }}\,{z^3} + \bar z + 198 = $

If ${z_1}{\rm{ and }}{z_2}$ be complex numbers such that ${z_1} \ne {z_2}$ and $|{z_1}|\, = \,|{z_2}|$. If ${z_1}$ has positive real part and ${z_2}$ has negative imaginary part, then $\frac{{({z_1} + {z_2})}}{{({z_1} - {z_2})}}$may be

The conjugate of a complex number is $\frac{1}{{i - 1}}$ then that complex number is

If $\frac{{z - \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and $\left| z \right| = 2$, then a value of $\alpha $ is

If $z$ is a complex number, then $(\overline {{z^{ - 1}}} )(\overline z ) = $