Properties of Complex Numbers
Properties of Complex Numbers
Properties of Complex Numbers
1. Addition: z 1+ z2 =( x 1 , y 1 ) + ( x 2 , y 2 )= ( x1 + x 2 , y 1 + y 2 )
2. Multiplication: z 1 z 2=( x 1 , y1 ) ( x 2 , y 2 ) =( x 1 x 2− y 1 y 2 , x 1 y 2 + y 1 x 2 )
3. Equality: ( x 1 , y 1 ) =( x 2 , y 2 ) iff. x 1=x 2 and y 1= y 2 .
1. Commutativity of Addition
For all z 1 , z 2 ∈C , z 1+ z2 =z2 + z 1. Let z 1=( x , y ) , z 2=( a , b ) . Then, by the definition of Addition
in Complex Number. We have,
z 1+ z2 =( x , y ) + ( a , b )
z 1+ z2 =( x+ a , y + b )
z 1+ z2 =( a+ x ,b + y ) → By Commutative Property of Addition
z 1+ z2 =( a , b ) + ( x , y ) → balik aninga form ( x1 , y 1 )+ ( x 2 , y 2 )
z 1+ z2 =z2 + z 1 .
2. Commutativity of Multiplication
3. Associativity of Addition
4. Associativity of Multiplication
z 1 ( z 2 z3 ) =[ ( xac− ybc )−( xbd + yad ) , ( xad − ybd ) + ( xbc+ yac ) ] → Combinelike terms
z 1 ( z 2 z3 ) =[ ( xa− yb ) c−( xb+ ya ) d , ( xa− yb ) d + ( xb+ ya ) c ] → factor out
z z z =[ xa− yb , xb + ya ] ( c , d ) → ilain si kabit
1 ( 2 3)
z 1 ( z 2 z3 ) =[ ( x , y ) ( a , b ) ] ( c , d ) → ibalik sa form ( x 1 , y 1 )( x2 , y 2 ) ibase ani ( x 1 x 2− y 1 y 2 , x 1 y 2 + y 1 x 2 )
z 1 ( z 2 z3 ) =( z 1 z 2 ) z 3 .
z 1 ( z 2 + z 3 )=z 1 z 2 + z 1 z 3 .
¿
Suppose there exists 0 and 0 such that for all z ∈ C
z +0=0+ z=z and z +0¿ =0¿ + z=z .
¿ ¿
Then 0 =0+ 0 =0.
¿
Hence, 0 =0.
8. Additive Inverse
z 1+ z2 =0 ⇔ ( x , y ) + ( a ,b )=(0,0)
⇔ ( x + a , y +b )=(0,0) →≝. of addition ∈complex number
⇔ x+ a=0 y +b=0
a=−x b=− y
Proof (Uniqueness)
To show that every z ∈ C has a unique additive inverse, we let – z =(−x ,− y ) and
z ¿ =( x ¿¿ ¿ , y ¿ ). ¿
¿ ¿
Suppose – z and z are additive inverses of z so that z + (−z )=0 and z + z =0. Then,
z ¿ =z¿ +0=(x ¿¿ ¿ , y ¿ )+(0,0) ¿
¿
¿( x ¿¿ ¿ , y )+( ( x , y )+(−x ,− y )) ¿ → z+ (−z ) =0
¿¿ → Associativity of addition
¿
¿ ( 0,0 ) +(−x ,− y ) → z+ z =0
¿ (−x ,− y ) → ( −x ,− y )=−z∧the inverse of −z is z .
¿z
¿
Therefore, z =z .
Subtraction
z 1−z 2=( x 1 , y 1 ) −( x 2 , y 2 )
9. Multiplicative Inverse
z 1 ∙ z 2=0 ⇔ ( x , y ) ∙ ( a , b ) =(1,0)
⇔ ( xa− yb , xb+ ya )=(1,0) →≝. of addition ∈complex number
xa− yb=1 xb+ ya=0
xa=1+ yb
xa 1+ yb
x
=
x [xb+ y ( 1+xyb )]=0
a=
1+ yb
x [ ( ) ]
xb+ y
1+ yb
x
=0 x
[ ( )]
2
−y 1
a= 1+ x 2 b+ y ( 1+ yb )=0
2
x +x y
2
x
Therefore, z 2= ( x +x y , x −+ xy y )=z
2 2 2 2
−1
1 .
−1
We denote the multiplicative inverse of z as z 1 for all z ∈C .
Proof: (Uniqueness)
To show that every z ∈ C has a unique multiplicative inverse: there exist z ' and z−1 such that
−1
z' ≠ z .
Since z ' is a multiplicative inverse,
z ' z=1 , where z ≠ 0 .
Hence,
' '
z =1∙ z
z =( z z ) z '
' −1
z =z ( z z )
' −1 ' → Associative Property
' −1
z =z ∙ 1
z ' =z−1 .
QED
Division:
z1 −1
For all z 1 , z 2 ∈ C , =z 1 z 2 where z 2 ≠ 0. If z 1=( x 1 , y 1 ) ∧z2 =( x 2 , y 2 ) . Then,
z2
z1 −1
=z z
z2 1 2
z1
z2 ( x2 − y2
= ( x 1 , y1 ) 2 2 , 2 2
x2+ y2 x2+ y2 )
→≝. of Multiplication∈complex number
z2 (
z1 x 1 x2 + y 1 y 2 y 1 x2− x1 y 2
=
x 22 + y 22
,
x 22 + y 22
.
)
QED
1
1. 2
=z−1
2 , where z 2 ≠ 0 .
z
Proof:
1 −1
For all z 2 ∈ C , 2
=z 2 where z 2 ≠ 0 .
z
By division we have,
z1 −1
=z 1 z 2 .
z2
1
=( 1 ) ( z 2 ) =z 2 .
−1 −1
Therefore, it follows that 2
z
( )
z1 1
2. =z 1 , where z 2 ≠ 0 .
z2 z2
Proof:
( )
z1 1
For all z 1 , z 2 ∈ C , =z 1 where z 2 ≠ 0 .
z2 z2
By division, we have,
z1 −1
=z z .
z2 1 2
−1 1
By property 1, we have shown that z 2 = .
z2
( )
z1 1
Therefore, =z 1 .
z2 z2
3. ( )( )
1
z1
1
=
1
z2 z1 z2
where ( z 1 ≠ 0 , z 2 ≠ 0 ) .
Proof:
By property 1, we have
( z1 )( z1 )=z
1 2
−1 −1
1 2 z .
−1 1
By property 1, ( z 1 z 2 ) = .
z1 z2
Therefore, ( )( )
1
z1
1
=
1
z2 z1 z2
.
z1 + z 2 z 1 z 2
4. = +
z3 z3 z3
Proof:
z1 + z 2 z 1 z 2
For all z 1 , z 2 , z 3 ∈C , = + .
z3 z3 z3
( )
z1 + z 2 1
By Property (2), =( z 1+ z2 ) .
z3 z3
z1 + z 2
=( z 1+ z2 ) ( z 3 )
−1
By Property (1),
z3
z1 + z 2
=z 1 z 3−1 + z 2 z 3−1
z3
( ) ( )
z1 + z 2 1 1
=z 1 + z2 .
z3 z3 z3
z1 + z 2 z 1 z 2
By Property (2), therefore, = + .
z3 z3 z3
Exercises.
1. Show that
(−1 ) z=−z
Solution.
Since z + (−1 ) z =z ( 1+ (−1 ) ) =z ( 0 )=0, then(−1 ) z=−z .
1
=z , ( z ≠ 0 )
1
z
Solution.
1 1 z z
= ∙ = =z , ( z ≠ 0 ) .
1 z −1 z 1
z
couple : ( z 1 z 2 ) ( z 3 z 4 )
¿ z1 [ z2 ( z3 z4 ) ] madi chin gibyaan ni padi Rommel nagpaduol sa ikaduhang couple
¿ z1 [ z3 ( z2 z4 ) ] hinay2x nalang ug palayo si smith, din nagpaka happy si Rommel kauban si lfred
¿ ( z1 z3 ) ( z2 z4 ). Guol na kaau si smith, unya gi comfort nalang syang madi chin, munang nagkadayon
nalang sila and stay strong si Alfred ug Rommel.
commute
Therefore , kung dili gani mag work ang relationship , pwdelang mag = perfect couple .
exchange
3. ( )( )
z1
z3
z2 z1 z2
= , ( z ≠ 0 , z 4≠ 0)
z 4 z3 z4 3
Solution.
( )( ) ( ) ( ) ( )( )
z1 z2
z3 z 4
=z 1
1
z3
z2
1
z4
=z 1 z 2
1
z3
1
z4
=z 1 z 2
1
( )
z z
= 1 2 , ( z3 ≠ 0 , z4 ≠ 0 ) .
z3 z4 z3 z4
4. Use number 3. To derive the cancellation law:
z1 z z 1
= , ( z ≠ 0 , z ≠ 0)
z2 z z 2 2
Solution.
z1 z z 1 z
=
z2 z z 2 z ( )( ) ( ) ( ) ( )
=
z1
z2
z
1
z
=
z1
z2
z
z2
z
z2 ( )
( z z−1 )= 1 1= 1 , ( z 2 ≠ 0 , z ≠ 0 ) .
Keywords
1 −1
Property 1: 2
=z 2 naa syay inverse
z
( )
z1 1
Property 2: =z 1 , naa syay whole number and fraction
z2 z2
Property 3: ( z )( z ) z z
1 1
1
=
1
2 1 2
kuyugon
z1 + z 2 z 1 z 2
Property 4: = + bulagon
z3 z3 z3
Denotei=( 0,1 ) .
So, i 2=¿ ( 0,1 ) ∙ ( 0,1 ) =(−1,0 )=−1 . By definition of multiplication in Complex Number
¿ x+ iy∨¿
¿ x+ yi
Addition: ( x 1 +iy 1 ) + ( x 2 +i y 2 ) =( x 1+ x 2 ) +i ( y 1+ y2 )
Division:
x1 +iy 1
x2 +i y 2
=
(
x1 x 2+ y 1 y 2 y 1 x2 −x1 y 2
2
x2+ y2
2
+i 2 2
x2+ y2 )
Definition 3:
Let z=x +iy ∈ Z . x is called the real part of z and y is called the imaginary part of z and we write
If ℜ ( z )=0 and ℑ( z)≠ 0 , then z is said to be a pure imaginary number. The conjugate of z is given by
z=x−iy .
Theorem
Let z , w ∈C . Then,
1 1
1. ℜ ( z )= ( z + z )and ℑ ( z )= ( z−z ).
2 2i
Proof:
Show that ℜ ( z )=x ,and let Show that ℑ ( z )= y , and let
z=x +iy . Then we have, z=x +iy . Then we have,
2x 2iy
¿ ¿
2 2i
¿x ¿y
¿ ℜ( z ) ¿ ℑ( z )
2. z +w=z +w
Proof:
Let z=x +iy and w=a+ ib. Then
z +w=( x+ iy ) + ( a+ib ) Substitute
¿ ( x+ a ) +i ( y+ b ) Lainon ang real and imaginary part
¿ ( x+ a )−i ( y +b ) conjugate
¿ ( x−iy ) + ( a−ib ) Usahon ang imaginary ug real part
¿ x+ iy+a+ ib conjugate
¿ z +w
z−w=z −w
Proof:
Let z=x +iy and w=a+ ib. Then
z−w=( x +iy )−( a+ib ) Substitute
¿ ( x−a )+ i ( y−b ) Lainon ang real and imaginary part
¿ ( x−a )−i ( y −b ) conjugate
¿ ( x−iy ) −( a−ib ) Usahon ang imaginary ug real
¿ x+ iy−a+ib part
conjugate
¿ z−w
3. zw=z w
Proof:
Let z=x +iy andw=a+ ib. Then
zw=( x+ iy ) ( a+ib )
¿ ( xa− yb ) +i ( xb +ay ) multiplication
¿ ( xa− yb )−i ( xb+ ay )
¿ ( x−iy ) ( a−ib )
¿ ( x+ iy ) ( a+ib )
¿zw
4. ( wz )= wz
Proof:
Let z=x +iy andw=a+ ib.
( x 1 x 2+ y 1 y 2 y 1 x 2−x 1 y 2
x22 + y 22
+i
x 22 + y 22 )
( wz )= xa+
a +b
yb ya−xb
+i
2 2
a +b
2 2
xa + yb ya−xb
¿2 2
−i 2 2
a +b a +b
xa yb iya ixb
¿ 2 2 + 2 2− 2 2 + 2 2
a +b a + b a +b a +b
xa +ixb iya− yb
¿ 2 2 − 2 2
a +b a +b
¿
|z 1 ± z 2|≤|z 1|+|z 2|
Proof: