The equation arg((z-1)/(z+1)) = π/4 represents a circle with: (1) centre at (0, -1) and radius √2 (2

Q: Wait, why is the locus a circle and not a straight line?

Great question! In Euclidean geometry, the locus of points that subtend a constant angle at two fixed points is a circular arc. If the angle were $\pi$ or $0$, it would be a line segment, but for any other constant angle, the set of points forms a part of a circle. We are essentially looking at the inscribed angle theorem in reverse!

Q: Why does the argument property $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$ hold?

This is a fundamental property of complex numbers derived from their polar form. If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, then $\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$. The argument is the exponent of $e$, which is $\theta_1 - \theta_2$. It works just like logarithms!

Q: How do we know the center is $(0, 1)$ and not $(0, -1)$?

After simplifying the equation to $x^2 + y^2 - 2y - 1 = 0$, we complete the square for $y$. We get $x^2 + (y^2 - 2y + 1) = 1 + 1$, which simplifies to $x^2 + (y-1)^2 = 2$. Comparing this to the standard circle equation $(x-h)^2 + (y-k)^2 = r^2$, we identify the center $(h, k)$ as $(0, 1)$.

Question

The equation $\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$ represents a circle with:

Options:

(A) centre at $(0, -1)$ and radius $\sqrt{2}$

(B) centre at $(0, 1)$ and radius $\sqrt{2}$

(C) centre at $(0, 0)$ and radius $\sqrt{2}$

(D) centre at $(0, 1)$ and radius 2

Sigma Percentile · Accepted Answer

### The Geometric Soul of Complex Numbers

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful problem involving complex numbers.

Often, students see $\arg\left(\frac{z-1}{z+1}ight) = \frac{\pi}{4}$ and immediately reach for complex algebraic manipulation. While that works, there is a deeper, more elegant geometric story hidden here.

Imagine you are standing on the complex plane, looking at two fixed points: $1$ and $-1$. The equation tells us that for any point $z$ on our locus, the angle formed by the lines connecting $z$ to $1$ and $z$ to $-1$ is exactly $\frac{\pi}{4}$.

This is a classic setup for the **inscribed angle theorem**. The locus of points subtending a constant angle at two fixed points is always a circular arc!

### The Algebraic Bridge

Now, let's translate this geometric intuition into the language of algebra. We set $z = x + iy$.

Our equation becomes $\arg(x - 1 + iy) - \arg(x + 1 + iy) = \frac{\pi}{4}$. This step is crucial.

By using the property $\arg\left(\frac{z_1}{z_2}ight) = \arg(z_1) - \arg(z_2)$, we have successfully broken down a complex quotient into a manageable difference of two arguments.

Now, recall that for any complex number $w = a + ib$, the argument is $	an^{-1}\left(\frac{b}{a}ight)$. Applying this to our terms, we get:

$$	an^{-1}\left(\frac{y}{x-1}ight) - 	an^{-1}\left(\frac{y}{x+1}ight) = \frac{\pi}{4}$$

### The Trigonometric Dance

This is where many students stumble, but you won't. We use the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$.

Let $A = \frac{y}{x-1}$ and $B = \frac{y}{x+1}$. Plugging these into our identity, we get:

$$	an^{-1}\left(\frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \left(\frac{y}{x-1}ight)\left(\frac{y}{x+1}ight)}ight) = \frac{\pi}{4}$$

Now, take the tangent of both sides. Since $	an\left(\frac{\pi}{4}ight) = 1$, the equation simplifies significantly.

The numerator becomes:

$$\frac{y(x+1) - y(x-1)}{(x-1)(x+1)} = \frac{2y}{x^2-1}$$

The denominator becomes:

$$1 + \frac{y^2}{x^2-1} = \frac{x^2 - 1 + y^2}{x^2 - 1}$$

### The Final Reveal

Watch the magic happen! When we divide the numerator by the denominator, the $(x^2-1)$ terms cancel out beautifully, leaving us with:

$$\frac{2y}{x^2 + y^2 - 1} = 1$$

Rearranging this, we get $x^2 + y^2 - 2y - 1 = 0$.

To find the center and radius, we complete the square for $y$:

$$x^2 + (y^2 - 2y + 1) = 1 + 1$$

This gives us the final equation:

$$x^2 + (y-1)^2 = 2$$

Comparing this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we find the **center is $(0, 1)$** and the **radius is $\sqrt{2}$**. You have just conquered the geometry of complex numbers!

Animated Solution for Mathematics - Complex Numbers: The equation $ar g (\frac{z - 1}{z + 1}) = \frac{π}{4}$ represents a circle with:

Select Answer:

Visualized Solution

Visualizing the Complex Plane

Geometric Intuition

Cartesian Substitution

Converting to Tangent Inverse

Applying Trigonometric Identity

Substitution into Formula

Simplifying the Numerator

Simplifying the Denominator

Forming the Equation

Completing the Square

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometric Soul of Complex Numbers

The Algebraic Bridge

The Trigonometric Dance

The Final Reveal

Similar Questions

The equation $∣ z - i ∣ = ∣ z - 1∣$ , $i = - 1$ , represents:

Find the centre and radius of circle given by $∣ \frac{z - α}{z - β} ∣ = k, k \neq = 1$ where $z = x + i y, α = α_{1} + i α_{2}, β = β_{1} + i β_{2}$ .

If $∣ z^{2} - 1∣ = ∣ z ∣^{2} + 1$ , then $z$ lies on

If $R e (\frac{z - 1}{2 z + i}) = 1$ , where $z = x + i y$ , then the point $(x, y)$ lies on a

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Let $z$ be a complex number such that $∣ \frac{z - 2 i}{z + i} ∣ = 2, z \neq = - i$ . Then $z$ lies on the circle of radius 2 and centre

If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies

If $ω = \frac{z}{z - \frac{1}{3} i}$ and $∣ ω ∣ = 1$ , then $z$ lies on

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation arg(z+1z−1​)=4π​ represents a circle with: