For any complex number w = c + id, let arg(w) ∈ (-π, π], where i = √-1. Let α and β be real numbers

Q: Wait, why is the locus of $\arg\left(\frac{z - z_1}{z - z_2}\right) = \theta$ an arc of a circle?

This is a classic result in complex geometry! The expression $\arg\left(\frac{z - z_1}{z - z_2}\right)$ represents the angle between the vectors $(z - z_2)$ and $(z - z_1)$. If this angle is constant, the locus of $z$ is an arc of a circle passing through $z_1$ and $z_2$. Think of it as the inscribed angle theorem from high school geometry, just extended to the complex plane.

Q: How do I know for sure that the endpoints are $-\alpha$ and $-\beta$?

Look at the form $\arg\left(\frac{z - (- \alpha)}{z - (- \beta)} ight)$. The standard form for the locus is $\arg\left(\frac{z - z_1}{z - z_2} ight) = \theta$, where $z_1$ and $z_2$ are the endpoints of the chord. By comparing the two, it is clear that $z_1 = -\alpha$ and $z_2 = -\beta$. Since $\alpha$ and $\beta$ are real, these points must lie on the real axis.

Q: Why did we need a test point? Couldn't we just solve for $\alpha$ and $\beta$ directly?

Great question! Solving the quadratic equation $x^2 + 5x + 4 = 0$ gives us the set $\{-\alpha, -\beta\} = \{-1, -4\}$. This tells us that the set $\{\alpha, \beta\}$ is $\{1, 4\}$. However, it does not tell us which is which. The argument condition $\arg\left(\frac{z + \alpha}{z + \beta}\right) = \frac{\pi}{4}$ is sensitive to the order of $\alpha$ and $\beta$. A test point is the most reliable way to break this symmetry and confirm the correct assignment.

Question

For any complex number $w = c + id$, let $\arg(w) \in (-\pi, \pi]$, where $i = \sqrt{-1}$. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers $z = x + iy$ satisfying $\arg\left(\frac{z + \alpha}{z + \beta}\right) = \frac{\pi}{4}$, the ordered pair $(x, y)$ lies on the circle $x^2 + y^2 + 5x - 3y + 4 = 0$. Then which of the following statements is (are) TRUE ?

Options:

(A) $\alpha = -1$

(B) $\alpha\beta = 4$

(C) $\alpha\beta = -4$

(D) $\beta = 4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the complex plane! Today, we are going to unravel a beautiful mystery hidden within the geometry of complex numbers.

Often, when we see an expression like $\arg\left(\frac{z + \alpha}{z + \beta}\right) = \frac{\pi}{4}$, it can look intimidating. But let us pause and look at it through the lens of geometry.

This equation is not just a random collection of symbols; it is a geometric signature. It tells us that the point $z$ moves along an arc of a circle, where the angle subtended by the chord connecting the points $-\alpha$ and $-\beta$ is constant at $\frac{\pi}{4}$.

This is a direct application of the inscribed angle theorem, beautifully translated into the language of complex numbers.

### The Real Axis Intersection: The Algebraic Key

Now, let us find where this circle touches the real axis. Since we are told that $\alpha$ and $\beta$ are real numbers, the points $-\alpha$ and $-\beta$ must lie on the x-axis, where the imaginary part is zero.

This is our golden opportunity! We are given the equation of the circle:
$$x^2 + y^2 + 5x - 3y + 4 = 0$$

To find the intersection points on the x-axis, we simply set $y=0$. The equation transforms into a simple quadratic:
$$x^2 + 5x + 4 = 0$$

Factoring this, we get $(x + 1)(x + 4) = 0$, which gives us the roots $x = -1$ and $x = -4$. These are our endpoints!

This means the set $\{-\alpha, -\beta\}$ is $\{-1, -4\}$, which implies that the set $\{\alpha, \beta\}$ is $\{1, 4\}$.

### The Verification: The Power of the Test Point

We know the set of values for $\alpha$ and $\beta$, but we have a final hurdle: which is which? Is $\alpha = 1$ and $\beta = 4$, or is it the other way around?

To find out, we need to test our assumption. Let us pick a point $z$ on the circle. If we set $x = -1$ in our circle equation, we get $y^2 - 3y = 0$, which gives us $y = 0$ or $y = 3$.

We already know $y=0$ is an endpoint, so let us choose the point $z = -1 + 3i$. Now, let us test the assumption $\alpha = 1$ and $\beta = 4$.

We evaluate the ratio $\frac{z + \alpha}{z + \beta}$ at $z = -1 + 3i$:
$$\frac{(-1 + 3i) + 1}{(-1 + 3i) + 4} = \frac{3i}{3 + 3i} = \frac{i}{1 + i}$$

To find the argument, we use the property $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$. Thus:
$$\arg\left(\frac{i}{1 + i}\right) = \arg(i) - \arg(1 + i) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$

It matches perfectly! The elegance of this result is undeniable.

By combining the geometric interpretation of the argument with the algebraic power of quadratic roots and a simple test point, we have unlocked the solution. The values are **$\alpha = 1$** and **$\beta = 4$**.

Select Answer:

Visualized Solution

Locus of the Argument

Endpoints on Real Axis

Intersection with Real Axis

Solving for Intercepts

Finding $- α$ and $- β$

Product of $α$ and $β$

Choosing a Test Point

Coordinates of the Test Point

Defining the Test Point

Testing $α = 1, β = 4$

Evaluating the Argument

Final Angle Calculation

Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Real Axis Intersection

The Verification

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