Question

If the locus of the complex number $z$ given by $arg(z+i)-arg(z-i)=\frac{2\pi}{3}$ is an arc of a circle, then the length of the arc is

Answer 1

$\frac{4\pi\sqrt{3}}{9}$

Answer 2

The given equation is $arg(z+i)-arg(z-i)=\frac{2\pi}{3}$.

Using the property $arg(z_1) - arg(z_2) = arg\left(\frac{z_1}{z_2}\right)$, we can rewrite the equation as: $arg\left(\frac{z+i}{z-i}\right) = \frac{2\pi}{3}$

Let $P_1$ be the point representing the complex number $i$, and $P_2$ be the point representing the complex number $-i$. Let $Z$ be the point representing the complex number $z$. The term $z-i$ represents the vector $\vec{P_1Z}$. The term $z+i$ represents the vector $\vec{P_2Z}$. The expression $arg\left(\frac{z+i}{z-i}\right)$ represents the angle from the vector $\vec{P_1Z}$ to the vector $\vec{P_2Z}$. This angle is $\angle P_1 Z P_2$. So, $\angle P_1 Z P_2 = \frac{2\pi}{3}$.

The points $P_1$ and $P_2$ are fixed points in the complex plane: $P_1=(0,1)$ and $P_2=(0,-1)$. The locus of a point $Z$ such that the angle $\angle P_1 Z P_2$ is constant is an arc of a circle passing through $P_1$ and $P_2$.

Let $R$ be the radius of this circle and $C$ be its center. The chord connecting $P_1$ and $P_2$ has length $d = |i - (-i)| = |2i| = 2$. The angle subtended by this chord at any point $Z$ on the arc is $\theta = \frac{2\pi}{3}$.

Since $\theta = \frac{2\pi}{3} = 120^\circ$, which is greater than $90^\circ$, the locus is the minor arc of the circle. The angle subtended by the chord $P_1P_2$ at the center $C$ of the circle, let's call it $\alpha$, is related to $\theta$ by $\alpha = 2(\pi - \theta)$. So, $\alpha = 2\left(\pi - \frac{2\pi}{3}\right) = 2\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}$.

Now we need to find the radius $R$ of the circle. Consider the triangle $\triangle P_1 C P_2$. It is an isosceles triangle with $CP_1 = CP_2 = R$. The angle $\angle P_1 C P_2 = \alpha = \frac{2\pi}{3}$. Let $M$ be the midpoint of the chord $P_1P_2$. $M=(0,0)$. The length $MP_1 = \frac{1}{2} |P_1P_2| = \frac{1}{2}(2) = 1$. In the right-angled triangle $\triangle CMP_1$, we have $\angle MCP_1 = \frac{\alpha}{2} = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}$. Using trigonometry: $\sin(\angle MCP_1) = \frac{MP_1}{CP_1}$ $\sin\left(\frac{\pi}{3}\right) = \frac{1}{R}$ $\frac{\sqrt{3}}{2} = \frac{1}{R}$ $R = \frac{2}{\sqrt{3}}$

The length of the arc is given by the formula $L = R \alpha$, where $\alpha$ is the central angle in radians corresponding to the arc. $L = \frac{2}{\sqrt{3}} \times \frac{2\pi}{3} = \frac{4\pi}{3\sqrt{3}}$ To rationalize the denominator, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $L = \frac{4\pi\sqrt{3}}{3\sqrt{3}\sqrt{3}} = \frac{4\pi\sqrt{3}}{9}$