Lecture 3:Moduli and ConjugatesDan SloughterFurman UniversityMathematics 39March 10, 20043.1 The modulus of a complex numberDefinition 3.1. For z = x + iy ∈ C, we call|z| =px2+ y2the modulus, or absolute value, of z.Note that if z = x + iy is real, that is, if y = 0, then|z| =√x2= |x|.That is, the modulus of a real number is just the ordinary absolute value.Geometrically, |z| is the distance between z and 0, the origin.Example 3.1. If z = 3 + i, then |z| =√9 + 1 =√10.Note that if z1= x1+ iy1and z2= x2+ iy2, then|z1− z2| =p(x1− x2)2+ (y1− y2)2is the distance between z1and z2.Example 3.2. It follows that the setC = {z ∈ C : |z + 2 − i| = 5}is a circle of radius 5 with center at −2 + i.1Also note that, since|z|2= (Re z)2+ (Im z)2,we have (Re z)2≤ |z|2and (Im z)2≤ |z|2, and hence|Re z| ≤ |z| and |Im z| ≤ |z|.3.2 Two inequalitiesWe may restate the familiar triangle inequality for points in the plane usingthe modulus of complex numbers:Proposition 3.1. For any complex numbers z1and z2,|z1+ z2| ≤ |z1| + |z2|.Note that, since | − z| = |z|,|z1| = |(z1+ z2) − z2| ≤ |z1+ z2| + |z2|,and so|z1+ z2| ≥ |z1| − |z2|.Similarly,|z1+ z2| ≥ |z2| − |z1|.Hence we have:Proposition 3.2. For any complex numbers z1and z2,|z1+ z1| ≥ ||z1| − |z2||.Example 3.3. Suppose z lies on the circle |z − i| = 1. Then, for example,|z − 1| = |(z − i) + (i − 1)| ≤ |z − i| + |i − 1| = 1 +√2and|z − 1| = |(z − i) + (i − 1)| ≥ ||z − i| − |i − 1|| = |1 −√2| =√2 − 1.23.3 ConjugatesDefinition 3.2. If z = x + iy ∈ C, we call¯z = x − iythe conjugate of z.Geometrically, ¯z is the reflection of z about the real axis. Note that¯¯z = zand|¯z| = |z|.It is easy to show thatz1+ z2= ¯z1+ ¯z2,z1− z2= ¯z1− ¯z2,z1z2= ¯z1¯z2,andz1z2=¯z1¯z2.Now if z = x + iy, thenz + ¯z = (x + iy) + (x − iy) = 2x,soRe z =z + ¯z2.Similarly,z − ¯z = (x + iy) − (x − iy) = 2iy,soIm z =z − ¯z2i.We also see thatz¯z = (x + iy)(x −iy ) = x2+ y2= |z|2.3Proposition 3.3. For any complex numbers z1and z2,|z1z2| = |z1||z2|and, if z26= 0,z1z2=|z1||z2|.Proof. The result follows from the f act that|z1z2|2= (z1z2)(z1z2) = (z1¯z1)(z2¯z2) = |z1|2|z2|2= (|z1||z2|)2.Example 3.4. For example, if |z| < 1, it follows that|3z2+ 4z − 2| ≤ 3|z|2+ 4|z| + 2 < 3 + 4 + 2 =