如何求满足arg((z-z₁)/(overline{z}-z₂))=π/4的复数z的轨迹？

Hey there! Let's work through this complex number trajectory problem together—no need to slog through messy x+iy expansions if we use some clever complex number properties and algebraic tricks.

Approach 1: Leverage Argument Properties to Simplify the Equation #

First, recall that for any non-zero complex numbers (A) and (B), arg(A/B) = arg(A) - arg(B). Also, notice that (\overline{z} - z_2 = \overline{z - \overline{z_2}}) (you can verify this by expanding (z_2 = c+id) and (\overline{z_2} = c-id)). Since arg(\overline{X}) = -arg(X) for any non-zero complex (X), we can rewrite the original condition as:

arg(z - z₁) + arg(z - \overline{z₂}) = π/4

Alternatively, since the argument is (\pi/4), we can express the ratio as a positive real multiple of (e^{i\pi/4}) (the complex number with argument (\pi/4) and positive modulus):

(z - z₁)/(overline{z} - z₂) = t·e^{i\pi/4}, where t > 0

Rearranging gives:

z - z₁ = t·e^{i\pi/4}·(overline{z} - z₂)

Approach 2: Eliminate the Parameter t to Get the Trajectory Equation #

To get rid of the positive real parameter (t):

1. Take the conjugate of both sides of the equation above:

overline{z} - \overline{z₁} = t·e^{-i\pi/4}·(z - \overline{z₂})

2. Solve for (t) from the first equation: (t = (z-z₁)/(e^{i\pi/4}(overline{z}-z₂))), then substitute this into the conjugated equation. After simplifying (using (e^{i\pi/4}·e{-i\pi/4}=1) and (e^{i\pi/2}=i)), we get:

i·(overline{z} - \overline{z₁})(overline{z} - z₂) = (z - z₁)(z - \overline{z₂})

3. Now substitute (z=x+iy), (z₁=a+ib), (z₂=c+id) into this equation, expand all terms, and combine like terms. You'll end up with a quadratic curve equation:

x² - 2xy - y² + (b - a - c - d)x + (a + b + c - d)y + (ac + ad - bc + bd) = 0

What Type of Curve Is This? #

To identify the curve, use the discriminant of the quadratic equation (Ax²+Bxy+Cy²+...=0):

• Here, (A=1), (B=-2), (C=-1)
• Calculate (B²-4AC = (-2)^2 - 4×1×(-1) = 4 + 4 = 8 > 0)

Since the discriminant is positive, this is a hyperbola. For the rotation angle: we use (\cot2θ=(A-C)/B=(1 - (-1))/(-2)=-1), which gives (2θ=3π/4) (or (θ=3π/8)), or equivalently a rotation of (-π/8)—this aligns with your initial guess of a hyperbola rotated by (\pi/8), just with a direction difference.

Quick Geometric Interpretation #

Geometrically, this trajectory represents all points (z) where the counterclockwise angle between the vector (z-z₁) and the vector (\overline{z}-z₂) is exactly (\pi/4). This fits the geometric definition of a hyperbola as the set of points with a constant angle relationship to two fixed points.

内容的提问来源于stack exchange，提问作者otreblig