Let arg(z) represent the principal argument of the complex number z. The, |z| = 3 and arg(z - 1) - a

Q: Wait, why does $\arg(z - 1) - \arg(z + 1) = \frac{\pi}{4}$ represent an arc and not a full circle?

Excellent observation! The argument function is multi-valued, but the principal argument is restricted to $(-\pi, \pi]$. When we fix the difference to a specific positive constant like $\frac{\pi}{4}$, we are selecting only one side of the chord connecting $1$ and $-1$. If it were just the magnitude, it would be a full circle, but the directed nature of the argument forces it to be a specific arc.

Q: How do I know the center of the arc lies on the imaginary axis?

Symmetry is your best friend in complex geometry! The points $1$ and $-1$ are symmetric with respect to the imaginary axis. Since the condition $\arg(z - 1) - \arg(z + 1) = \frac{\pi}{4}$ treats $1$ and $-1$ symmetrically, the resulting locus must also be symmetric about the imaginary axis. Therefore, the center must lie on the $y$-axis.

Q: Can I solve this using the algebraic form $z = x + iy$?

You absolutely can, but it is a path of thorns! Substituting $z = x + iy$ into $\tan^{-1}(\dots) - \tan^{-1}(\dots) = \frac{\pi}{4}$ leads to a messy algebraic equation. While it works, it obscures the beautiful geometric intuition that the JEE examiners want you to see. Geometry is the 'shortcut' to elegance here.

Q: Why did we calculate the maximum distance from the origin to check for intersection?

Think of it as a 'reach' test. If the entire arc is contained within a circle of radius $2.414$, and our target circle has a radius of $3$, the arc is essentially 'trapped' inside the larger circle. If the maximum reach of the inner shape is less than the radius of the outer shape, they can never meet.

Q: Is it possible for the arc to intersect the circle at two points if the radius was different?

Yes, absolutely! If the radius of the circle $|z| = R$ were smaller, say $R = 2$, the circle would cut through the arc at two distinct points. The beauty of this problem lies in the specific values chosen, which force the arc to stay strictly inside the circle.

Question

Let $\arg(z)$ represent the principal argument of the complex number $z$. The, $|z| = 3$ and $\arg(z - 1) - \arg(z + 1) = \frac{\pi}{4}$ intersect:

Options:

(A) Exactly at one point

(B) Exactly at two points

(C) Nowhere

(D) At infinitely many points

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the complex plane. Today, we are mapping a landscape to find the intersection of two distinct mathematical entities: a circle and an arc.

We are standing at the origin of the complex plane, constrained by two specific rules. Let us uncover them one by one.

### The Guardian Circle

Our first constraint is foundational: $|z| = 3$. In the language of geometry, this represents the set of all points $z$ whose distance from the origin is exactly $3$.

This is a circle centered at the origin $(0, 0)$ with a radius $R_1 = 3$. This serves as our primary boundary.

### The Arc of Constant Angle

Now, consider the second condition: $\arg(z - 1) - \arg(z + 1) = \frac{\pi}{4}$. Using the property $\arg(A) - \arg(B) = \arg(\frac{A}{B})$, we simplify the expression:

$$\arg\left(\frac{z - 1}{z + 1}ight) = \frac{\pi}{4}$$

This is a classic result in complex analysis. The locus of points $z$ such that the angle subtended by the segment connecting $z_1 = 1$ and $z_2 = -1$ is a constant $	heta$ is an arc of a circle.

Since $	heta = \frac{\pi}{4}$, which is positive and less than $\pi$, we are dealing with a major arc. It connects the points $(1, 0)$ and $(-1, 0)$.

### Finding the Heart of the Arc

To understand the geometry of this arc, we must determine its center and radius. We know the angle at the circumference is $\frac{\pi}{4}$, so the angle subtended at the center must be double that: $2 	imes \frac{\pi}{4} = \frac{\pi}{2}$.

Because the points $1$ and $-1$ are symmetric about the imaginary axis, the center $C$ of this arc must lie on the imaginary axis at some point $(0, k)$. Consider the triangle formed by the center $(0, k)$, the origin $(0, 0)$, and the point $(1, 0)$.

This is a right-angled triangle where the distance from the center to $(1, 0)$ is the radius $R$. By the Pythagorean theorem:

$$R^2 + R^2 = (1 - (-1))^2$$

$$2R^2 = 4 \implies R^2 = 2 \implies R = \sqrt{2}$$

Since the distance from $(0, k)$ to $(1, 0)$ is $\sqrt{2}$, we have $k^2 + 1^2 = (\sqrt{2})^2$, which yields $k = 1$. Our arc is part of a circle centered at $(0, 1)$ with radius $\sqrt{2}$.

### The Final Confrontation

We now compare our two shapes. The first is a circle of radius $3$ centered at the origin. The second is an arc of a circle centered at $(0, 1)$ with radius $\sqrt{2}$.

To determine if they intersect, we find the point on the arc furthest from the origin. This point lies on the imaginary axis at a distance of $k + R$ from the origin:

$$y_{max} = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$$

Compare this to our first circle, which has a radius of $3$. Since $2.414 < 3$, the entire arc is contained strictly within the interior of the circle $|z| = 3$.

There is no overlap and no intersection. The two paths never cross. Through pure geometric reasoning, we have proven that these two conditions define worlds that never meet; the set of intersection points is **empty**.

Animated Solution for Mathematics - Complex Numbers: Let $ar g (z)$ represent the principal argument of the complex number $z$ . The, $∣ z ∣ = 3$ and $ar g (z - 1) - ar g (z + 1) = \frac{π}{4}$ intersect:

Select Answer:

Visualized Solution

Visualizing $∣ z ∣ = 3$

The Argument Condition

Locus of the Argument Equation

Finding the Center of the Arc

Calculating Radius and Center

Maximum Distance from Origin

Comparing the Two Loci

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Guardian Circle

The Arc of Constant Angle

Finding the Heart of the Arc

The Final Confrontation

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