Question

if z1=a+ib and z2=c+id are complex numbers such that |z1|=|z2|=1 and re(z1z2?)=0 then the pair of complex numbers w1=a+ic and w2=b+id satisfies

Answer 1

The question is incomplete. The complete question is as:

If $z_{1}$=a+ib and $z_{2}$=c+id are complex numbers such that |$z_{1}$|=|$z_{2}$|=1 and $Re(z_{1}\overline{z_{2}})=0$ then the pair of complex numbers $w_{1}$=a+ic and $w_{2}$=b+id satisfies

1. $\left | w_{1} \right |= 1$

2. $\left | w_{2} \right |= 1$

3. $Re(w_{1}\overline{w_{2}})=0$

4. None of these

Solution:

Since |$z_{1}$|=|$z_{2}$|=1

So, $z_{1} = \cos \Theta_{1}+i \sin \Theta_{1}$

and $z_{2} = \cos \Theta_{2}+i \sin \Theta_{2}$

So, $a= \cos \Theta_{1} , b= \sin \Theta_{1}, c = \cos \Theta_{2}, d= \sin \Theta_{2}$

Consider $Re(z_{1}\overline{z_{2}})$

= $(\cos \Theta _{1}+i\sin \Theta _{1})(\cos \Theta _{2}-i\sin \Theta _{2})$

= $=(\cos \Theta _{1}\cos \Theta _{2}-i\cos \Theta _{1}\sin \Theta _{2}+i\sin \Theta_{1}\cos \Theta _{2}-i^{2}\sin \Theta _{1}\sin \Theta _{2})$

= $=(\cos \Theta _{1}\cos \Theta _{2}+\sin \Theta _{1}\sin \Theta _{2})+i(\sin \Theta _{1}\cos \Theta _{2}-\cos \Theta_{1}\sin \Theta _{2})$

= $=\cos (\Theta _{1}-\Theta _{2})+i\sin (\Theta _{1}-\Theta _{2})$

Since $Re(z_{1}\overline{z_{2}}) =0$

So, $\cos (\Theta _{1}-\Theta _{2})=0$

$(\Theta _{1}-\Theta _{2})=\frac{\Pi }{2}$

So, $\Theta _{1}=\Theta _{2}+\frac{\Pi }{2}$

Now, consider $w_{1}=a+ic$

$w_{1}=\cos \Theta _{1}+i\cos \Theta _{2}$

$w_{1}=\cos (\frac{\Pi }{2}+\Theta _{2})+i\cos \Theta _{2}$

$w_{1}=\sin \Theta _{2}+i\cos \Theta _{2}$

$\left |w_{1} \right |=\sqrt{(\sin \Theta _{2})^{2}+(\cos \Theta _{2})^{2}}=1$

So, $\left |w_{1} \right |=1$

$w_{2}=\cos \Theta _{2}+i\sin \Theta _{2}$

$\left |w_{2} \right |=\sqrt{(\sin \Theta _{2})^{2}+(\cos \Theta _{2})^{2}}=1$

So, $\left |w_{2} \right |=1$

Consider

$Re(w_{1}\overline{w_{2}})$

=$Re[(\cos \Theta _{1}+i\cos \Theta _{2})(\sin \Theta _{1}-i\sin \Theta _{2})]$

$=[(\cos \Theta _{1}\sin \Theta _{1}+\cos \Theta _{2}\sin \Theta _{2})]$

=$[(\cos (\frac{\Pi }{2}+\Theta _{2})\sin (\frac{\Pi }{2}+\Theta _{2})+\cos \Theta _{2}\sin \Theta _{2})]$

=$-\sin \Theta _{2}\cos\Theta _{2}+\cos \Theta _{2}\sin \Theta _{2}$

= 0.

So, $Re(w_{1}\overline{w_{2}})=0$

So, the options 1, 2 and 3 are correct.