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

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(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 ${

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(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$

List $I$	List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$	$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\| \ldots . .\left\|1-z_9\right\|}{10}$ equals	$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	$4.$ $2$

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

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