A point z moves in the complex plane such that arg((z-2)/(z+2)) = π/4, then the minimum value of |z

Q: Wait, why is the locus only an arc and not the full circle?

Great observation! The equation $\arg\left(\frac{z-z_1}{z-z_2}\right) = \theta$ defines a specific arc because the angle $\theta$ is fixed. If $\theta = \frac{\pi}{4}$, the point $z$ must maintain that specific angle relative to the chord connecting $2$ and $-2$. If it were the full circle, the angle would change as you move to the other side of the chord.

Q: Why can't we just use the distance formula directly on the complex number?

You could, but it would be incredibly messy! By converting the complex locus into a Cartesian circle equation, we transform a difficult algebraic minimization problem into a simple geometric one: finding the shortest distance from a point to a circle, which is just the distance to the center minus the radius.

Q: Why do we subtract the radius $r$ from the distance $CP$?

Think of the geometry: the shortest path from any point $P$ outside a circle to the circle itself is along the line segment connecting $P$ to the center $C$. This segment passes through the circle at exactly one point. The length of this segment is the distance $CP$, and the distance from that point on the circle to $P$ is $CP - r$.

Question

A point $z$ moves in the complex plane such that $\arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{4}$, then the minimum value of $|z - 9\sqrt{2} - 2i|^2$ is equal to .

Sigma Percentile · Accepted Answer

### The Geometry of Complex Motion

Welcome, fellow traveler in the world of complex numbers! Today, we are going to unravel a problem that might look like a daunting algebraic beast, but is actually a beautiful piece of geometry in disguise.

We are given a point $z$ moving in the complex plane such that $\arg\left(\frac{z-2}{z+2}ight) = \frac{\pi}{4}$. Our mission is to find the minimum value of $|z - 9\sqrt{2} - 2i|^2$.

Let us embark on this journey together.

### Phase 1: Decoding the Locus

First, let us look at the condition $\arg\left(\frac{z-2}{z+2}ight) = \frac{\pi}{4}$. In the complex plane, the argument of a quotient represents the angle subtended by the segment connecting the two points in the denominator and numerator.

Specifically, this equation tells us that the angle subtended by the chord connecting $z_1 = 2$ and $z_2 = -2$ at the point $z$ is constant at $\frac{\pi}{4}$. This is the definition of an arc of a circle.

### Phase 2: The Algebraic Transformation

Let $z = x + iy$. Our equation becomes:

$$\arg(x - 2 + iy) - \arg(x + 2 + iy) = \frac{\pi}{4}$$

Recalling that $\arg(x+iy) = 	an^{-1}\left(\frac{y}{x}ight)$, we can write this as:

$$	an^{-1}\left(\frac{y}{x-2}ight) - 	an^{-1}\left(\frac{y}{x+2}ight) = \frac{\pi}{4}$$

Now, we apply the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$. Taking the tangent of both sides, we get:

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

### Phase 3: Unveiling the Circle

Let us simplify this expression carefully. The numerator becomes:

$$\frac{y(x+2) - y(x-2)}{(x-2)(x+2)} = \frac{4y}{x^2-4}$$

The denominator becomes:

$$1 + \frac{y^2}{x^2-4} = \frac{x^2-4+y^2}{x^2-4}$$

When we divide these, the $(x^2-4)$ terms cancel out, leaving us with $\frac{4y}{x^2+y^2-4} = 1$. Rearranging this, we obtain $x^2 + y^2 - 4y - 4 = 0$.

By completing the square for $y$, we see:

$$x^2 + (y^2 - 4y + 4) = 4 + 4 \Rightarrow x^2 + (y-2)^2 = 8$$

This is a circle centered at $C = (0, 2)$ with a radius $r = \sqrt{8} = 2\sqrt{2}$.

### Phase 4: The Final Minimization

We need to minimize $|z - (9\sqrt{2} + 2i)|^2$. This is the square of the distance between our moving point $z$ and the fixed point $P = (9\sqrt{2}, 2)$.

The distance $CP$ between the center $(0, 2)$ and the point $P(9\sqrt{2}, 2)$ is simply $9\sqrt{2}$. Since $P$ lies outside the circle, the minimum distance to the circle is:

$$d_{min} = CP - r = 9\sqrt{2} - 2\sqrt{2} = 7\sqrt{2}$$

Finally, the square of this distance is:

$$(7\sqrt{2})^2 = 49 	imes 2 = \mathbf{98}$$

We have arrived at our destination. The beauty of this problem lies in how the complex algebra melts away to reveal a simple, elegant geometric truth.

Animated Solution for Mathematics - Complex Numbers: A point $z$ moves in the complex plane such that $ar g (\frac{z - 2}{z + 2}) = \frac{π}{4}$ , then the minimum value of $∣ z - 92 - 2 i ∣^{2}$ is equal to .

Enter Numerical Value:

Visualized Solution

Understanding the Locus

Applying Argument Properties

Converting to Cartesian Form

Applying $tan^{- 1}$ Formula

Simplifying the Equation

Finding the Circle's Equation

Identifying Center and Radius

Identifying the Target Point $P$

Calculating Distance $C P$

Finding Minimum Distance

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Motion

Phase 1

Phase 2

Phase 3

Phase 4

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