Question

Let $z=x+iy$ and $w=u+iv$ be complex numbers on the unit circle such that $z^2+w^2=1$. Then the number of ordered pairs $(z,w)$ is

• A. $0$
• B. $4$
• C. $8$
• D. infinite

Answer 1

The correct option is C $8$

Assuming the two complex numbers as
$z=e^{i\alpha }$ and $w=e^{i\beta }$
$z^2+w^2=1\Rightarrow e^{2i\alpha }+e^{2i\beta }=1\Rightarrow \cos 2\alpha +i\sin 2\alpha +\cos 2\beta +i\sin 2\beta =1$
$\cos 2\alpha +\cos 2\beta =1$.............(i)
and
$\sin 2\alpha +\sin 2\beta =0$
We know that
$\sin 2\alpha =-\sin 2\beta \Rightarrow {\sin }^{2}2\alpha ={\sin }^{2}2\beta \Rightarrow 1-{\cos }^{2}2\alpha =1-{\cos }^{2}2\beta \Rightarrow \cos 2\alpha =\pm \cos 2\beta$
$\cos 2\alpha =-\cos 2\beta$ not possible.
So, $\cos 2\alpha =+\cos 2\beta$
Put in equation (i)
$2\cos 2\alpha =1\Rightarrow \cos 2\alpha =\frac{1}{2};\cos 2\beta =\frac{1}{2}2\alpha =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}$ and
$2\beta =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}$
From the condition
$\sin 2\alpha =-\sin 2\beta$
If $2\alpha$ is in first quadrant than $2\beta$ should be in fourth quadrant and vice versa.
So if we choose $2\alpha =\frac{\pi }{3}$ then it will have two possible values of $2\beta =\frac{5\pi }{3},\frac{11\pi }{3}$
Similarly for other values
Hence the total numbers of order pairs are $=8$