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Concept 02 / 10Complex Numbers and Quadratic Equations

Geometry of Multiplication: Rotation, the Triangle Inequality and Useful Modulus Identities

Multiplication by e^(i theta) as rotation, the rotation formula for vectors about any point, the triangle inequality and its equality cases, and identities such as the parallelogram law used for extremum and locus problems.

Figure 1
Since moduli multiply and arguments add, multiplying zz by reiθre^{i\theta} scales its length by rr and turns it anticlockwise through θ\theta. In particular, multiplying by eiθe^{i\theta} is a pure rotation, and multiplying by i=eiπ/2i = e^{i\pi/2} is a quarter turn. This is the real reason complex numbers are so effective in plane geometry. Rotation about a general point. To rotate the point z1z_{1} about the point z0z_{0} through angle θ\theta, rotate the vector z1−z0z_{1} - z_{0} and then add back z0z_{0}: the image is z0+(z1−z0)eiθz_{0} + (z_{1} - z_{0})e^{i\theta}. More generally, if the segment z0z2z_{0}z_{2} is obtained from the segment z0z1z_{0}z_{1} by turning through θ\theta anticlockwise, then z2−z0z1−z0=∣z2−z0∣∣z1−z0∣eiθ\frac{z_{2} - z_{0}}{z_{1} - z_{0}} = \frac{|z_{2} - z_{0}|}{|z_{1} - z_{0}|}e^{i\theta}. The figure shows this setup. Everything about the angle at z0z_{0} is encoded in the quotient. The argument of z2−z0z1−z0\frac{z_{2} - z_{0}}{z_{1} - z_{0}} is the angle from z0z1z_{0}z_{1} to z0z2z_{0}z_{2}, measured anticlockwise. Consequences: the three points are collinear if and only if this quotient is real, and z0z1⊥z0z2z_{0}z_{1} \perp z_{0}z_{2} if and only if it is purely imaginary. An equilateral triangle z1z2z3z_{1}z_{2}z_{3} is characterised by z12+z22+z32=z1z2+z2z3+z3z1z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1}, which comes from applying the 60∘60^{\circ} rotation to two sides. The triangle inequality. For any complex numbers, ∣z1+z2∣≤∣z1∣+∣z2∣|z_{1} + z_{2}| \le |z_{1}| + |z_{2}|. Geometrically, one side of a triangle is at most the sum of the other two. Equality holds exactly when z1z_{1} and z2z_{2} point in the same direction, that is, when one is a non-negative real multiple of the other (arg⁡z1=arg⁡z2\arg z_{1} = \arg z_{2} when both are non-zero). The companion inequality is ∣z1−z2∣≥∣∣z1∣−∣z2∣∣|z_{1} - z_{2}| \ge ||z_{1}| - |z_{2}||, with equality when they point in the same direction. Together: ∣∣z1∣−∣z2∣∣≤∣z1±z2∣≤∣z1∣+∣z2∣||z_{1}| - |z_{2}|| \le |z_{1} \pm z_{2}| \le |z_{1}| + |z_{2}|. These are the main tool for maximum and minimum modulus problems. For instance, if ∣z−3−4i∣=2|z - 3 - 4i| = 2, then zz lies on a circle of radius 22 about 3+4i3 + 4i, which is at distance 55 from the origin. So ∣z∣|z| ranges over [5−2,5+2]=[3,7][5 - 2, 5 + 2] = [3, 7], with the extremes on the line through the origin and the centre. Modulus identities worth knowing. ∣z∣2=zzˉ|z|^{2} = z\bar{z}, which converts modulus conditions into algebra. ∣z1+z2∣2=∣z1∣2+∣z2∣2+2Re(z1zˉ2)|z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + 2\text{Re}(z_{1}\bar{z}_{2}), the complex form of the cosine rule. Parallelogram law: ∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)|z_{1} + z_{2}|^{2} + |z_{1} - z_{2}|^{2} = 2(|z_{1}|^{2} + |z_{2}|^{2}), meaning the squares of the diagonals of a parallelogram add up to the squares of the four sides. ∣z1+z2∣=∣z1−z2∣  ⟺  Re(z1zˉ2)=0  ⟺  |z_{1} + z_{2}| = |z_{1} - z_{2}| \iff \text{Re}(z_{1}\bar{z}_{2}) = 0 \iff the vectors are perpendicular (the parallelogram is a rectangle, so its diagonals are equal). If ∣z∣=1|z| = 1 then zˉ=1z\bar{z} = \frac{1}{z}, which is extremely handy when points lie on the unit circle. Key Takeaways: - Multiplying by reiθre^{i\theta} scales by rr and rotates by θ\theta; multiplying by ii rotates by 90∘90^{\circ}. - Rotation of z1z_{1} about z0z_{0} by θ\theta: z0+(z1−z0)eiθz_{0} + (z_{1} - z_{0})e^{i\theta}. - arg⁡z2−z0z1−z0\arg\frac{z_{2} - z_{0}}{z_{1} - z_{0}} is the angle from z0z1z_{0}z_{1} to z0z2z_{0}z_{2}; real means collinear, purely imaginary means perpendicular. - ∣∣z1∣−∣z2∣∣≤∣z1+z2∣≤∣z1∣+∣z2∣||z_{1}| - |z_{2}|| \le |z_{1} + z_{2}| \le |z_{1}| + |z_{2}|. - Equality in the triangle inequality holds when the vectors are parallel and point the same way. - Parallelogram law: ∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)|z_{1} + z_{2}|^{2} + |z_{1} - z_{2}|^{2} = 2(|z_{1}|^{2} + |z_{2}|^{2}). - On the unit circle, zˉ=1z\bar{z} = \frac{1}{z}. - Pitfall: rotating about the origin when the question asks for rotation about another point. Brain-Teaser 1: Two vertices of an equilateral triangle are 00 and 3+i33 + i\sqrt{3}. Find the third vertex. Solution: Rotate the vector from 00 to z1=3+i3z_{1} = 3 + i\sqrt{3} through ±60∘\pm 60^{\circ} about 00. With eiπ/3=12+i32e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}: z1eiπ/3=(3+i3)(12+32i)=32+332i+32i−32=23iz_{1}e^{i\pi/3} = (3 + i\sqrt{3})(\frac{1}{2} + \frac{\sqrt{3}}{2}i) = \frac{3}{2} + \frac{3\sqrt{3}}{2}i + \frac{\sqrt{3}}{2}i - \frac{3}{2} = 2\sqrt{3}i. With e−iπ/3e^{-i\pi/3}: (3+i3)(12−32i)=32−332i+32i+32=3−3i(3 + i\sqrt{3})(\frac{1}{2} - \frac{\sqrt{3}}{2}i) = \frac{3}{2} - \frac{3\sqrt{3}}{2}i + \frac{\sqrt{3}}{2}i + \frac{3}{2} = 3 - \sqrt{3}i. So the third vertex is 23i2\sqrt{3}i or 3−3i3 - \sqrt{3}i. Both check out: ∣z1∣=12|z_{1}| = \sqrt{12} and each candidate is at distance 12\sqrt{12} from both 00 and z1z_{1}. Rotation gives both answers directly, one for each orientation, without solving any simultaneous equations. Brain-Teaser 2: Find the greatest and least values of ∣z∣|z| if ∣z−4z∣=2|z - \frac{4}{z}| = 2. Solution: Let r=∣z∣r = |z|. By the triangle inequality, ∣z∣−∣4z∣≤∣z−4z∣|z| - |\frac{4}{z}| \le |z - \frac{4}{z}|, so r−4r≤2r - \frac{4}{r} \le 2, giving r2−2r−4≤0r^{2} - 2r - 4 \le 0 and r≤1+5r \le 1 + \sqrt{5}. Similarly 4r−r≤2\frac{4}{r} - r \le 2 gives r2+2r−4≥0r^{2} + 2r - 4 \ge 0 and r≥5−1r \ge \sqrt{5} - 1. Both bounds are attained on the real axis: for real z=r>0z = r > 0, the condition reads ∣r−4r∣=2|r - \frac{4}{r}| = 2, which gives r=1+5r = 1 + \sqrt{5} (since 41+5=5−1\frac{4}{1 + \sqrt{5}} = \sqrt{5} - 1) or r=5−1r = \sqrt{5} - 1 (since 45−1=5+1\frac{4}{\sqrt{5} - 1} = \sqrt{5} + 1). So the maximum is 5+1\sqrt{5} + 1 and the minimum is 5−1\sqrt{5} - 1. The key idea is to bound ∣z∣|z| using the two-sided triangle inequality, then check that the bounds can actually be achieved. Brain-Teaser 3: z1,z2,z3z_{1}, z_{2}, z_{3} lie on the unit circle and z1+z2+z3=0z_{1} + z_{2} + z_{3} = 0. Show that they form an equilateral triangle. Solution: The centroid of the triangle is z1+z2+z33=0\frac{z_{1} + z_{2} + z_{3}}{3} = 0, which is also the circumcentre, because all three points are at distance 11 from 00. A triangle whose centroid and circumcentre coincide is equilateral, since each median is then perpendicular to the opposite side. An algebraic check: ∣z1−z2∣2=2−2Re(z1zˉ2)|z_{1} - z_{2}|^{2} = 2 - 2\text{Re}(z_{1}\bar{z}_{2}). From z3=−(z1+z2)z_{3} = -(z_{1} + z_{2}) we get 1=∣z1+z2∣2=2+2Re(z1zˉ2)1 = |z_{1} + z_{2}|^{2} = 2 + 2\text{Re}(z_{1}\bar{z}_{2}), so Re(z1zˉ2)=−12\text{Re}(z_{1}\bar{z}_{2}) = -\frac{1}{2} and ∣z1−z2∣2=3|z_{1} - z_{2}|^{2} = 3. By symmetry every side is 3\sqrt{3}. Complex numbers turn 'centroid equals circumcentre' into a single line of algebra.

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Geometry of Multiplication: Rotation, the Triangle Inequality and Useful Modulus Identities — DPP

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