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Three distinct points $x, y, z$ lie on a unit circle of the complex plane and satisfy $x+y+z=0$. Then $x, y, z$ form the vertices of .

1. An isosceles but not equilateral triangle.
2. An equilateral triangle.
3. A triangle of any shape.
4. A triangle whose shape can't be determined.
5. None of the above.

$|x| = |y| = |z| = 1$ because points are on the unit circle.

Suppose, points $x,y$ $\text{and}$ $z$ makes angles $\theta_1,\theta_2$ $\text{and}$ $\theta_3$ respectively from +ve x-axis in anticlockwise direction.

$\therefore x = e^{i\theta_1}, y = e^{i\theta_2}, z = e^{i\theta_3}$

$x+y+z = 0$

$\frac{x+y+z}{z} = \frac{0}{z},z \neq 0 \Rightarrow \frac{x+y+z}{z} = 0$

$1+\frac{x}{z} + \frac{y}{z} = 0 \Rightarrow \frac{x}{z} + \frac{y}{z} = -1 \Rightarrow \left |\frac{x+y}{z} \right| = |-1| \Rightarrow \frac{|x+y|}{|z|}= 1 \Rightarrow |x+y| = 1$

$|e^{i\theta_1} + e^{i\theta_2}| = 1 \Rightarrow (\cos \theta_1 + \cos\theta_2)^2 + (\sin \theta_1 + \sin\theta_2)^2 = 1$

$\Rightarrow \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 = \frac{-1}{2} \Rightarrow \cos(\theta_1 – \theta_2) = \frac{-1}{2}$

For $\theta \in [0, 360^{\circ}), |\theta_1 – \theta_2| = 120^{\circ}\; or\; |\theta_1 – \theta_2| = 240^{\circ}\;$

So, there are $2$ possibilities if we measure angle anticlockwise.

In the left figure, these 2 possibilities are shown for point $y$. The angle between lines $ox$ and $oy$ is $120^{\circ}$ and $240^{\circ}$.

We do the same thing for point $z$, so, divide by $y$ both sides in $x+y+z=0$ and do the same thing as above and we get the middle figure. Now, out of $2$ possibilities for each $y$ and $z$, if we fix $y$ then the other will be $z$, and suppose, we get the figure which is on the right side.

In the figure which is on the right side, since $ox=oy$, so angle between lines $oy$ & $yx$ and lines $ox$ & $yx$ will be same and it will be of $30^{\circ}$ because angle between $ox$ and $oy$ is $120^{\circ}$.

If we do the same for triangles $yoz$ and $zox$, we get all $3$ angles  $\measuredangle zyx, \measuredangle yxz, \measuredangle xzy = 60^{\circ}.$

$\text{Hence,}$ $\textbf{Option (B)}$

Since the arrows must be symmetric to each other as none of them is speacial. So equilateral.

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