Conjugate. The conjugate of z=x+iy is z=x−iy — obtained by flipping the sign of the imaginary part, equivalently by reflecting z across the real axis in the Argand plane. A key fact: the product of a complex number with its own conjugate is always a non-negative real number, zz=(x+iy)(x−iy)=x2+y2.
Ten conjugate properties (each provable directly from the definition, several proved in the text):
z1+z2=z1+z2
z1−z2=z1−z2
z1z2=z1z2
(z2z1)=z2z1,z2=0
Re(z)=2z+z
Im(z)=2iz−z
zn=(z)n, n an integer
z is real⟺z=z
z is purely imaginary⟺z=−z
z=z
Proof idea (property 1): writing z1=x1+iy1,z2=x2+iy2, z1+z2=(x1+x2)−i(y1+y2)=(x1−iy1)+(x2−iy2)=z1+z2. Proof idea (property 9): z=−z⟺x+iy=−(x−iy)=−x+iy⟺2x=0⟺x=0, i.e. z is purely imaginary.
The conjugate is the standard tool for dividing by a complex number: multiplying numerator and denominator by the conjugate of the denominator makes the denominator real (exactly like rationalising a surd).
Modulus. The modulus of z=x+iy, written ∣z∣, is ∣z∣=x2+y2 — the distance from z to the origin in the Argand plane, generalising the real-number absolute value. Note zz=∣z∣2.
Using ∣zk∣2=zkzk turns each conjugate into a reciprocal-like expression, and combining z1+z2+z3 over a common denominator z1z2z3 reproduces exactly the numerator given in the problem.
Step 1. Convert each modulus condition into a conjugate identity.
Since ∣z1∣=1: z1z1=1⇒z1=z11.
Since ∣z2∣=2: z2z2=4⇒z2=z24.
Since ∣z3∣=3: z3z3=9⇒z3=z39.
Step 2. Add the three conjugates over the common denominator z1z2z3.
Dividing 12 by ∣z1∣+∣z2∣+∣z3∣=6 (sum instead of product) — coincidentally the same denominator here, but conceptually wrong and fails on similar problems with different moduli …