Question about triangle formed by points lying on unit circle in Argand Plane following specific condition.

Question

If $\mathrm{A}\left({z}_1\right), \mathrm{B}\left({z}_2\right)$ and $\mathrm{C}\left({z}_3\right)$ be vertices of a $\triangle \mathrm{ABC}$ such that $\left|{z}_1\right|=\left|{z}_2\right|=\left|{z}_3\right|=1$ and there exist $\alpha \in\left(0, \frac{\pi}{2}\right)$ such that $z_1+z_2 \cos \alpha+z_3 \sin \alpha=0$, then what is the value of $\bar{z}_2 z_3+z_2 \bar{z}_3$?
My attempt
The magnitude of all the numbers is equal to $1$ so obviously they lie in on the unit circle but I don't really know how to interpret the second condition. I tried thinking that somehow I could take projections of $z_2$ and $z_3$ onto a line opposite to $z_1$ to make $-z_1$(like vectors) but I didn't get anywhere with that.
PS:I encountered this problem in my institute's practice sheets for an undergraduate college entrance exam (JEE). So it would be preferable if the answers didn't use very advanced theorems that are not possible to understand at my level.

Answer 1

One way to bring the mentioned second condition into play is to express all given complex numbers in the form, $z=\cos\left(\theta\right)+i\sin\left(\theta\right)$.

Therefore, we have, $$z_1=\cos\left(\theta_1\right)+i\sin\left(\theta_1\right),\quad\space z_2=\cos\left(\theta_2\right)+i\sin\left(\theta_2\right), \quad\space\text{and}\quad\space z_3=\cos\left(\theta_3\right)+i\sin\left(\theta_3\right).$$

Now, we can express the second condition as shown below. $$\cos\left(\theta_1\right)+\cos\left(\alpha\right)\cos\left(\theta_2\right) +\sin\left(\alpha\right)\cos\left(\theta_3\right)+i\left(\sin\left(\theta_1\right)+ \cos\left(\alpha\right)\sin\left(\theta_2\right)+ \sin\left(\alpha\right)\sin\left(\theta_3\right)\right)=0.$$

Therefore, we have, $$\cos\left(\theta_1\right)=-\cos\left(\alpha\right)\cos\left(\theta_2\right)-\sin\left(\alpha\right)\cos\left(\theta_3\right)=0\qquad\text{and}\tag{1}$$ $$\sin\left(\theta_1\right)=- \cos\left(\alpha\right)\sin\left(\theta_2\right)- \sin\left(\alpha\right)\sin\left(\theta_3\right)=0.\qquad\quad\enspace\tag{2}$$

If we square (1) and (2), and then add those two together, we get, $$\small{\cos^2\left(\theta_1\right)+\sin^2\left(\theta_1\right)= \cos^2\left(\alpha\right)+ \sin^2\left(\alpha\right)+2\cos\left(\alpha\right)\sin\left(\alpha\right)\left(\cos\left(\theta_2\right)\cos\left(\theta_3\right)+ \sin\left(\theta_2\right)\sin\left(\theta_3\right)\right)}.$$

Once simplified, this becomes, $$1=1+\sin\left(2\alpha\right)\cos\left(\theta_2-\theta_3\right)$$

From this, we can deduce that, $$\cos\left(\theta_2-\theta_3\right)=0.\tag{3}$$

Now, let us express $\bar{z}_2z_3$ and $z_2\bar{z}_3$. $$\bar{z}_2z_3=\left(\cos\left(\theta_2\right)+i\sin\left(\theta_2\right)\right) \left(\cos\left(\theta_3\right)-i\sin\left(\theta_3\right)\right)\tag{4}$$ $$z_2\bar{z}_3=\left(\cos\left(\theta_2\right)-i\sin\left(\theta_2\right)\right) \left(\cos\left(\theta_3\right)+i\sin\left(\theta_3\right)\right)\tag{5}$$

Using (4) and (5), we shall write, $$\bar{z}_2z_3+ z_2\bar{z}_3=2\left(\cos\left(\theta_2\right) \cos\left(\theta_3\right)+ \sin\left(\theta_2\right) \sin\left(\theta_3\right)\right)=2\cos\left(\theta_2-\theta_3\right)\tag{6}.$$ According to (3) and (6), $\bar{z}_2z_3+ z_2\bar{z}_3=0$.