Bisector of an imaginary angle may be real

Let's find the locus of points in R² defined by the homogeneous quadratic equation

(1)	AX² + BXY + CY² = 0.

with A, B, C real. Considering X/Y as a single variable x we see that (1) can be solved by a quadratic formula

(2)	X/Y = (-B ± √B² - 4AC) / 2A.

Denoting the two roots a = (-B + √B² - 4AC) / 2A and b = (-B - √B² - 4AC) / 2A we see that

(3)	AX² + BXY + CY² = A(X - aY)(X - bY) = 0.

Thus solving (1) is equivalent to solving (X - aY)(X - bY) = 0, or X - aY = 0 and X - bY = 0 separately. The latter equations define two lines through the origin. Since, depending on whether B² - 4AC is positive, zero, or negative a and b may be different and real, equal and real, or complex with a non-zero imaginary part, we may say that the two lines so defined are real and different, real and coincident, and imaginary.

For a straight line X - aY = 0 that passes through the origin, a is the tangent of the angle between the line and the Y-axis; b in X - bY is interpreted similarly. If the two angles are α and β, the angle (up to a sign) between the two lines is α - β. Since ab = C/A, we can find its tangent:

(4)	tan(α - β)	= (tan α - tan β) / (1 + tan α · tan β)
		= (a - b) / (1 + ab)
		= √B² - 4AC / (C + A).

The equation of the bisector between the two lines is X - mY = 0, with m interpreted similarly. The bisector forms an angle with the Y-axis which, if taken twice gives the sum of the angles of the two lines. Using the slopes this shows that

(5)	2m / (1 - m²) = (a + b) / (1 - ab) = - B / (A - C).

As an equation for m, (5) can be rewritten:

(6)	Bm² - 2(A - C) m - B = 0.

This is a quadratic equation in m with the discriminant equal to 4((A - C)² + B²) / B² which is always positive. It follows that (6) always has two real solutions: one gives the bisector of the (internal) angle formed by the two lines X - aY = 0 and X - bY = 0 and the other gives the bisector of the external angle. Observe that, since (from (6)) the product of the two roots (the slopes of the bisectors) is -1, the two bisectors are perpendicular, as one would expect.

Now the situation should give one a start. (1) defines a pair of lines. The lines, generally speaking, may be real or imaginary. In either case they appear to form an angle whose bisector is always a real line! What do we make out of this? Is the oddity real or imaginary?

The quandary admits a very simple and quite a natural answer. To visualize complex numbers that comprise two real quantities, one needs two real axes - a real plane. Since, in (1), both X and Y may come out complex, one may need two pairs of axes, i.e., a 4-dimensional space to visualize their mutual dependency. This is where the imaginary lines live. What we see in the real XY-plane is just a projection - often inadequate - of a 4-dimensional construct. From a 4D perspective, there is nothing strange in imaginary lines having real bisectors. This may happen even in 3D. Indeed, draw an angle and its bisector in the z = 0 plane (the xy-plane in 3D.) A rotation of the plane around the bisector line will move the two sides of the angle into the third dimension - invisible, so to speak, to the inhabitants of the plane - whereas the bisector itself will firmly remain in the plane.

By analogy, the explanation to the phenomenon of the imaginary angle that has a real bisector is that sitting in 3D we do not see the complete picture which is rather naturally 4D but only its cross-section. Sometimes, as in the case of an imaginary angle, the cross-section does not give a clue of what it is we are seeing. Sometimes, as in the case of a real angle, we get a better idea of a (complex) phenomenon. We may always wonder whether, in the latter case, we see a complete picture.

References

1. G. Salmon, Treatise on Conic Sections, Chelsea Pub, 6e, 1960

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