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.
Write z+6+8i as z−(−(6+8i)) and apply the reverse/direct triangle inequality ∣∣z1∣−∣z2∣∣≤∣z1+z2∣≤∣z1∣+∣z2∣ with z1=z and z2=6+8i.
Step 1. Compute the modulus of the constant. Let w=6+8i. Then ∣w∣=62+82=36+64=100=10.
Step 2. State the triangle-inequality bound. For any complex numbers z1,z2, ∣∣z1∣−∣z2∣∣≤∣z1+z2∣≤∣z1∣+∣z2∣ (properties (2)/(5) of §2.5.1). Apply with z1=z,z2=w=6+8i, so z1+z2=z+6+8i. …