If {z_1} = a + ib and {z_2} = c + id are comp | vedclass

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

Question

(d)Since $|{z_1}| = |{z_2}| = 1$, we have

${z_1} = \cos {	heta _1} + i\sin {	heta _1},{z_2} = \cos {	heta _2} + i\sin {	heta _2}$

where ${

Vedclass · Accepted Answer

All the above

Vedclass · Answer

(d)Since $|{z_1}| = |{z_2}| = 1$, we have

${z_1} = \cos {	heta _1} + i\sin {	heta _1},{z_2} = \cos {	heta _2} + i\sin {	heta _2}$

where ${	heta _1} = arg({z_1})$ and ${	heta _2} = arg({z_2})$

Also, ${z_1} = a + ib$ and ${z_2} = c + id.$

Therefore$a = \cos {	heta _1}$,$b = \sin {	heta _1},c = \cos {	heta _2}$ and $d = \sin {	heta _2}$

Also, $R({z_1}{\overline z _2}) = 0$

==> $R[(\cos {	heta _1} + i\sin {	heta _1})(\cos {	heta _2} - i\sin {	heta _2})] = 0$

==> $R[(\cos ({	heta _1} - {	heta _2}) + i\sin ({	heta _1} - {	heta _2})] = 0$

==> $\cos ({	heta _1} - {	heta _2}) = 0$

==> ${	heta _1} - {	heta _2} = \frac{\pi }{2}$

==> ${	heta _1} = {	heta _2} + \frac{\pi }{2}$

Now, ${w_1} = a + ic = \cos {	heta _1} + i\cos {	heta _2}$$ = \cos {	heta _1} + i\sin {	heta _1}$

==> $|{w_1}| = 1$

Similarly, $|{w_2}| = 1$

Next ${w_1}{\overline w _2} = (\cos {	heta _1} + i\sin {	heta _1})(\cos {	heta _2} - i\sin {	heta _2})$

$ = \cos ({	heta _1} - {	heta _2}) + i\sin ({	heta _1} - {	heta _2})$

==> $|{w_1}{\overline w _2}| = 1$

Finally, $R({\overline w _1}{w_2}) = R({w_2}{\overline w _1})$

$ = R[(\cos {	heta _2} + i\sin {	heta _2})(\cos {	heta _1} - i\sin {	heta _1})]$

$ = R[\cos ({	heta _2} - {	heta _1}) + i\sin ({	heta _2} - {	heta _1})]$

$ = \cos ({	heta _2} - {	heta _1}) = \cos \left( {\frac{{ - \pi }}{2}} ight) = 0$

If ${z_1} = a + ib$ and ${z_2} = c + id$ are complex numbers such that $|{z_1}| = |{z_2}| = 1$ and $R({z_1}\overline {{z_2}} ) = 0,$ then the pair of complex numbers ${w_1} = a + ic$ and ${w_2} = b + id$ satisfies

Similar Questions

Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$be any complex number, then $\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right| = $

If $z$ is a complex number, then $z.\,\overline z = 0$ if and only if

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

The modulus and amplitude of $\frac{{1 + 2i}}{{1 - {{(1 - i)}^2}}}$ are

If $|{z_1}|\, = \,|{z_2}|$ and $amp\,{z_1} + amp\,\,{z_2} = 0$, then