Lecture 4 Complex Analysis

Absolute value:-
The absolute value or modulus of Complex number z=a+ib is defined as
|z|=|a+ib|= ?(a^2+b^2 )
EX :- Find
|2-3i|
|4+5i| H.W
SOL:-
|2-3i|=?(4+9)=?13

Absolute Value Properties:-
|Z|=?(Z?Z ? )?|Z|^2=ZZ ?
|Z|=|Z ? |
|Z_1-Z_2 |=|Z_2-Z_1 |
1/2 (Z+Z ? )=a??(a^2+b^2 )=|z|?R(Z)?|Z|
1/2i (Z+Z ? )=b??(a^2+b^2 )=|z|?I(Z)?|Z|
|Z_1/Z_2 |=|Z_1 |/|Z_2 | ,|Z_1 Z_2 |=|Z_1 ||Z_2 |
To prove (6) , we will use property (1)
|Z_1/Z_2 |=?(Z_1/Z_2 ((Z_1/Z_2 ) ) ? )=?(Z_1/Z_2 ?Z ?_1/Z ?_2 )=?((Z_1 Z ?_1)/(Z_2 Z ?_2 ))=?((Z_1 Z ?_1)/(Z_2 Z ?_2 ))=?(Z_1 Z ?_1 )/?(Z_2 Z ?_2 )=|Z_1 |/|Z_2 |
|Z_1 Z_2 |=?((Z_1 Z_2 ) ((Z_1 Z_2 ) ) ? )=?(Z_1 Z_2 Z ?_1 Z ?_2 )=?(Z_1 Z ?_1 Z_2 Z ?_2 )=?(Z_1 Z ?_1 ) ?(Z_2 Z ?_2 )
=|Z_1 ||Z_2 |
Triangle Inequality:-
|Z_1+Z_2 |?|Z_1 |+|Z_2 |

Proof:- |Z_1+Z_2 |^2=(Z_1+Z_2 ) ((Z_1+Z_2 ) ) ?
=(Z_1+Z_2 )(Z ?_1+Z ?_2 )
=Z_1 Z ?_1+Z_1 Z ?_2+Z_2 Z ?_1+Z_2 Z ?_2

Let U=Z_1 Z ?_2 , then U ?= ((Z_1 Z ?_2 ) ) ?=Z ?_1 Z ?_2=Z ?_1 Z_2=Z_2 Z ?_1
Z_1 Z ?_2+Z_2 Z ?_1=U+U ?=2R(U)?2|U|=2|Z_1 Z ?_2 |=2|Z_1 ||Z ?_2 |=2|Z_1 ||Z_2 |
Thus
|Z_1+Z_2 |^2? Z_1 Z ?_1+2|Z_1 ||Z_2 |+Z_2 Z ?_2=|Z_1 |^2+2|Z_1 ||Z_2 |+|Z_2 |^2=(Z_1+Z_2 )^2
by root for two side, We get
|Z_1+Z_2 |?|Z_1 |+|Z_2 |
|Z_1-Z_2 |? |(|Z_1 |-|Z_2 | )|
Proof:-
|Z_1 |=|Z_1-Z_2+Z_2 |?|Z_1-Z_2 |+|Z_2 |
?|Z_1-Z_2 |?|Z_1 |-|Z_2 |
?|Z_1-Z_2 |? |(|Z_1 |-|Z_2 | )|
by take absolute value for two sides.