Animated Solution for Mathematics - Complex Numbers: The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

A Journey into Locus

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey to visualize the complex plane. Many students fear complex numbers because they treat them as abstract algebraic entities.

But in the JEE Advanced, the complex plane is a canvas. It is a place where geometry and algebra dance together. Let us decode this locus problem step by step.

Finding the Anchor

Every complex locus problem has a 'heart'—a point from which all other features emanate. Look closely at the provided Argand diagram. Do you see the point $A(-1,0)$?

Notice how the two rays originate from this point, and the dashed circular arc is centered exactly there. In the language of complex numbers, this point is $z_0=-1$.

When you see a diagram like this, your first instinct should be to shift your coordinate system. Instead of thinking about $z$ in isolation, think about $z$ relative to $z_0$.

We are interested in the vector $z-z_0$, which is $z-(-1)=z+1$. This simple shift is the key that unlocks the entire problem. Everything we do from here on will be in terms of $z+1$.

The Angular Dance

Now, let us look at the rays. A ray starting from a point $z_0$ at an angle $\theta$ is defined by the equation $\arg (z-z_0)=\theta$. In our diagram, we have two rays originating from $z_0=-1$.

The upper ray makes an angle of $\frac{\pi }{4}$ with the positive real direction, and the lower ray makes an angle of $-\frac{\pi }{4}$.

Since our shaded region is trapped between these two rays, the argument of our vector $(z+1)$ must be bounded by these two values. Mathematically, we write this as:

Do you see the elegance here? Because the region is symmetric about the real axis, we can condense this inequality. Using the property of absolute values, we can write this as:

This is our first condition. It tells us that the 'direction' of our complex number $z$ relative to $-1$ is restricted to a specific wedge.

The Radial Barrier

Next, we must address the circular boundary. A circle in the complex plane is defined by the set of points at a constant distance from a center. The equation is $|z-z_0|=r$.

We already know our center is $z_0=-1$. But what is the radius $r$?

Look at the diagram again. The arc passes through the point $C(1,0)$. The distance from the center $A(-1,0)$ to $C(1,0)$ is simply the difference in their real parts:

Thus, the radius of our circle is $r=2$.

Now, consider the shaded region. Is it inside the circle or outside? It is clearly extending outwards, away from the center.

This means the distance of any point $z$ in the shaded region from the center $A$ must be strictly greater than the radius. Therefore, our second condition is $|z-(-1)|>2$, or simply:

The Synthesis

We have now broken down the problem into two distinct, manageable constraints. We have the angular constraint, which restricts the 'direction' of $z$, and the radial constraint, which restricts the 'distance' of $z$.

To define the locus of $z$ that lies in the shaded region, we must satisfy both conditions simultaneously. We combine them with the logical 'and':

1. The radial condition: $|z+1|>2$ 2. The angular condition: $|\arg (z+1)|<\frac{\pi }{4}$

When you look at the options provided in the question, you will see that this combination perfectly matches the first option.

I want you to pause and appreciate what you just did. You didn't just solve an equation; you translated a visual image into the language of complex analysis.

You identified the anchor, you defined the angular boundaries, and you established the radial limit. This is the essence of JEE Advanced physics and mathematics—the ability to look at a complex system, break it down into its fundamental components, and reconstruct it using the laws of mathematics.

Keep this mindset, and no problem will ever be too intimidating. You are ready.