Animated Solution for Mathematics - Complex Numbers: Let $S_1=\{z\in \mathbb{C}:|z|\le 5\}$, $S_2=\{z\in \mathbb{C}:\text{Im}\left(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i}\right)\ge 0\}$ and $S_3=\{z\in \mathbb{C}:\text{Re}(z)\ge 0\}$. Then the area of the region $S_1\cap S_2\cap S_3$ is :

A Journey Through $S_1,S_2,$ and $S_3$

Welcome, fellow traveler of the complex plane! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of inequalities.

But I promise you, by the time we are done, you will see it for what it truly is: a beautiful, elegant slice of a circle. Let us embark on this journey together.

The Foundation ($S_1$ and $S_3$)

First, let us look at $S_1=\{z\in \mathbb{C}:|z|\le 5\}$. In the language of the Argand plane, $|z|$ is simply the distance of a point $z$ from the origin.

So, $S_1$ is the set of all points whose distance from the origin is at most $5$. Geometrically, this is a solid disk of radius $5$ centered at $(0,0)$. Imagine a target on a wall; $S_1$ is the entire bullseye.

Next, we have $S_3=\{z\in \mathbb{C}:\text{Re}(z)\ge 0\}$. If we write $z=x+iy$, then the real part is just $x$.

The condition $x\ge 0$ tells us that we are restricted to the right half of the complex plane. We have now sliced our disk in half, keeping only the part on the right side of the y-axis.

The Mystery of $S_2$

Now, let us confront the beast: $S_2=\{z\in \mathbb{C}:\text{Im}\left(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i}\right)\ge 0\}$. It looks intimidating, but let us breathe. We substitute $z=x+iy$ into the expression:

To find the imaginary part, we must rationalize. We multiply the numerator and denominator by the conjugate of the denominator, $1+\sqrt{3}i$. The denominator becomes $1^2+(\sqrt{3}{)}^2=4$.

The numerator becomes:

We only care about the imaginary part of this numerator. Extracting it, we get:

Since the denominator is a positive real number ($4$), the condition $\text{Im}(\text{Expression})\ge 0$ simplifies beautifully to $\sqrt{3}x+y\ge 0$, or $y\ge -\sqrt{3}x$. This is the region on or above the line $y=-\sqrt{3}x$.

The Intersection and the Final Area

We have our three constraints: 1. Inside the circle of radius $5$. 2. To the right of the y-axis ($x\ge 0$). 3. Above the line $y=-\sqrt{3}x$.

Visually, this is a circular sector. The upper boundary is the positive y-axis, which corresponds to an angle of $90^{\circ }$ or $\frac{\pi }{2}$ radians.

The lower boundary is the line $y=-\sqrt{3}x$. Since the slope is $-\sqrt{3}$, the angle $\theta$ satisfies $\tan \theta =-\sqrt{3}$, which gives $\theta =-60^{\circ }$ or $-\frac{\pi }{3}$ radians.

The total angle of our sector is the difference between these two boundaries:

Finally, the area of a circular sector is given by $A=\frac{1}{2}{r}^2\theta$. Substituting $r=5$ and $\theta =\frac{5\pi }{6}$:

And there you have it! Through simple geometry and a bit of algebraic housekeeping, we have found the area to be $\frac{125\pi }{12}$. Never fear the complex expression; just peel back the layers, and you will find the geometry waiting for you underneath.