Complex number: lecture 3
Complex number: lecture 3
Property of modulus
z=a+ibzˉ=a−ib
∣z∣:=a2+b2
(1) z=x+iy
⇒−∣z∣≤Re(z)≤∣z∣
- [Re(z)]2=x2≤x2+y2
⇒∣Re(z)∣≤x2+y2=∣z∣
Similarly, −∣z∣≤Im(z)≤∣z∣
Complex number: lecture 3
Property of modulus
(2) ∣z∣≥0for all z∈C
∣z∣=0iffz=0
Proof:
Consider ∣z∣2=0
⇔x2+y2=0
⇔x=0=y
⇔z=0
Complex number: lecture 3
Properties of modulus
(3) ∣z∣=∣−z∣=∣zˉ∣
(4) z.zˉ=∣z∣2
(5) ∣z1⋅z2∣=∣z1∣⋅∣z2∣
Complex number: lecture 3
Properties of modulus
Proof:
Consider ∣z1⋅z2∣2=(z1⋅z2)(z1⋅z2)
=(z1⋅z2)∣z1∣ˉ⋅∣z2∣ˉ
=∣z1∣2∣z2∣2
∣z1⋅z2∣=∣z1∣∣z2∣
Complex number: lecture 3
Triangle inequality
(6) ∣z1+z2∣≤∣z1∣+∣z2∣ ∀z1,z2∈C
Complex number: lecture 3
Proof
∣z1+z2∣2=(z1+z2)⋅(z1+z2){∵∣z∣2=z⋅zˉ}
=(z1+z2)(zˉ1+zˉ2)
=z1⋅zˉ1+z1zˉ2+z2zˉ1+z2⋅zˉ2
=∣z1∣2+z1zˉ2+z1zˉ2+∣z∣2
=∣z1∣2+2Re(z1zˉ2)+∣z2∣2
Complex number: lecture 3
Proof
∣z1+z2∣2=∣z1∣2+2Re(z1zˉ2)+∣z2∣2
We know that,
∣Re(z)∣≤∣z∣
≤∣z1∣2+2∣z1zˉ2∣+∣z2∣2
≤∣z1∣2+2∣z1∣⋅∣z2∣+∣z2∣2
≤(∣z1∣+∣z2∣)2
Thus,
∣z1+z2∣≤∣z1∣+∣z2∣
Complex number: lecture 3
Condition for equality
Question Is : ∣z1+z2∣=∣z1∣+∣z2∣?
- ∣z1+z2∣≤∣z1∣+∣z2∣ ⋯(1)
- The equality appears in (1) provided
- Re(z1zˉ2)=∣z1zˉ2∣
- This is equivalent to z1=t z2
- where t is a non negative real number.
Complex number: lecture 3
General inequality for complex number
∣z1+z2∣≤∣z1∣+∣z2∣ ⋯(1)
⇒∣z1∣−∣z2∣≤∣z1+z2∣
Consider ∣z1∣=∣z1+z2−z2∣
By (1), ∣z1∣≤∣z1+z2∣+∣z2∣
∣z1∣−∣z2∣≤∣z1+z2∣
Interchange z1 and z2, ∣z2∣−∣z1∣≤∣z1+z2∣
⇒∣∣z1∣−∣z2∣∣≤∣z1±z2∣≤∣z1∣+∣z2∣
Complex number: lecture 3
Property of modulus
(7) ∣z−1∣=∣z∣−1,0=z∈C.
Proof:
z⋅z1=1
∣z⋅z1∣=1
⇒∣z∣⋅∣z1∣=1
⇒∣z∣−1=∣z−1∣
Complex number: lecture 3
Property of modulus
(8) ∣z2z1∣=∣z2∣∣z1∣0=z2∈C
Proof:
∣z1⋅z21∣=∣z1∣⋅∣z2−1∣
=∣z1∣∣z2∣−1
=∣z2∣∣z1∣
Complex number: lecture 3
Parallelogram law
∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)
Proof:
L.H.S =∣z1+z2∣2+∣z1−z2∣2
=(z1+z2)(z1+z2)+(z1−z2)(z1−zˉ2)
Complex number: lecture 3
Proof
L.H.S. =∣z1+z2∣2+∣z1−z2∣2
=(z1+z2)(zˉ1+zˉ2)+(z1−z2)(zˉ1−zˉ2)
=∣z1∣2+∣z2∣2+z1zˉ2+zˉ1z2+∣z1∣2+∣z2∣2−z1zˉ2−zˉ1z2
=2(∣z1∣2+∣z2∣2)= R.H.S.
Complex number: lecture 3
Problem 1
Prove that if ∣z1∣=1=∣z2∣ andz1z2=−1, then 1+z1z2z1+z2 is a real number.
Solution:
A=1+z1z2z1+z2claim: A=Aˉ
Aˉ=(1+z1z2z1+z2)=1+zˉ1zˉ2zˉ1+zˉ2
Given
∣z1∣2=1⇒z1zˉ1=1⇒zˉ1=z11
Similarly, zˉ2=z21
Complex number: lecture 3
Solution
Aˉ=1+z11.z21z11+z21=1+z1z2z1+z2=A
⇒A is a real number.
Complex number: lecture 3
Remark
∣z∣=1 ⇒zzˉ=1 ⇒zˉ=z1
∣zˉ∣=∣z∣1=1.
Explanation:
Consider U={z∈C:∣z∣=1}
∣z∣=1
z=x+iy∣z∣2=x2+y2=1
∣z∣=1 then zˉ=z1
Complex number: lecture 3
Observation
(1)z1,z2∈U ⇒z1.z2∈U
(2)z∈U⇒z−1∈U,z−1=zˉ
Complex number: lecture 3
Theorem
Identity:
Let z1,z2,z3,z4∈C then ,
(z1−z2)(z3−z4)+(z1−z4)(z2−z3)=(z1−z3)(z2−z4)
Proof:
L.H.S =(z1−z2)(z3−z4)+(z1−z4)(z2−z3)
=z1z3−z1z4−z2z3+z2z4+z1z2−z1z3−z4z2+z4z3=z1z2−z1z4−z3z2+z3z4
R.H.S =(z1−z3)(z2−z4)
=z1z2−z1z4−z3z2+z3z4
L.H.S=R.H.S
Complex number: lecture 3
Problem 2
Show that if A,B,C,D are points in a plane then AD.BC≤BD.CA+CD.AB
Solution:
-
(z1−z4)(z2−z3)=(z1−z3)(z2−z4)−(z1−z2)(z3−z4)
-
∣z1−z4∣∣z2−z3∣≤∣z1−z3∣∣z2−z4∣+∣z1−z2∣∣z3−z4∣
-
AD.BC≤AC.BD+AB.CD
Complex number: lecture 3
Unimodular
z is said to be unimodular if ∣z∣=1.
Complex number: lecture 3
Problem 3
Let z1,z2∈C. Suppose 2−z1zˉ2z1−2z2is unimodular and z2 is not unimodular. Then, the point z1 lies on a
(a) straight line parallel to x-axis.
(b) straight line parallel to y-axis.
(c) circle of radius 2 (i.e., ∣z1∣=2)
(d) circle of radius 2.
Complex number: lecture 3
Solution
Given ∣2−z1zˉ2z1−2z2∣=1 & ∣z2∣=1
∣z1−2z2∣2=∣2−z1zˉ2∣2
⇒(z1−2z2)(z1−2z2)=(2−z1zˉ2)(2−z1zˉ2)
⇒(z1−2z2)(zˉ1−2zˉ2)=(2−z1zˉ2)(2−zˉ1z2)
⇒∣z1∣2+4∣z2∣2−2z1zˉ2−2zˉ1z2
=4+∣z1zˉ2∣2−2zˉ1z2−2z1zˉ2
⇒∣z1∣2+4∣z2∣2−4−∣z1zˉ2∣2=0
Complex number: lecture 3
Solution
⇒∣z1∣2(1−∣z2∣2)−4(1−∣z2∣2)=0
⇒(1−∣z2∣2)(∣z1∣2−4)=0
∣z2∣=1⇒∣z1∣2=4
⇒∣z1∣=2
Option (c) is correct.
