, Complex Number System
Solve for a :
x2 + 1 =
0
1
S
x2 =
x
=
=Fonti
↑
not real
Solve for a in :
d"-1 =
0
i) IR
x4 -
1 =
0
(x2+ 1)(x21) =
0 not iR
(x2 + 1)
(x + 1)(x 1)
-
=
0 ↑
x = = |[or = F]
ii)
x4 -
1 =
0
(x2+ 1)(x21) =
0 not iR
(x2 + 1)
(x + 1)(x 1)
-
=
0 ↑
x = = |[or = F]
iii) D (complex no . s
x4 -
1 =
0
(x2+ 1)(x[ 1) -
=
o
(x2 + 1)
1)(x 1) 0
=
(x +
-
x =
+
;t
iv) I
x4 -
1 =
0
(x2 1)(x[ 1)
+ -
=
o
(x2 + 1)
1)(x 1) 0
=
(x +
-
x =
1
Naming Complex Numbers
Symbols :
z
=
atbi
a + bi
1
real imaginary F
Ey : z
,
=
3 -
2)
z2
= 2+ gi
real part imaginery
↑ ↑
IRe (z , ) =
3 and Im(z ,) = -
2
IRe (zz) =
2 and
Im(zz) =
S
,Adding ↑ Subtracting Complex No S
.
1(z , zz + =
3 -
21 + 2 + 3i
=
5 + 3)
2 z
,
-
zz
=
(3 2i) (2 - si)
-
-
=
3 -
2i -
2 +
si
=
1 -
7
multiplying Complex No S .
3)z , xzy (3 2i)(z + gi) = -
4i -10 ;
2
=
b+ 1Si -
=
6 + 11) -
10) 1) -
=
6 + 11i + 10
=
10 + 11
in
Conjugate of z
If z = atbi
,
the conjugate,* =
a -
bi
Dividing Complex No S .
5) Z 3-2i
six2-si
=
~
↑ I
2+ 2 31
-
(3
-
-
zi)(z - si)
(2+ si) (z-si)
2
6-1Si-ui + ioi
?
4- loi + 10i 2si -
I
al
: Complex numbers are closed (ie : stay in the complex set) under +; -jx ; +
Argand Planes
x = Real
y Imaginary
=
Let z = 4 + 3)
im ·
(4 + 3i)
O
121
i
!
Y Re
U
length of z
modulus of z
= 121Y Argument of z : f
(a)
3
1 : =3 = Es tans-b 0 :
tan"
(i)
b
(z) =
5 cost =
1z) 0 =
cos"
| = an + b
=
sind i O sin"
:
(i)
, Modulus-Argument Form
z =
4 + 3i
z = (z) (cosf ising)+
-sinf =
ilzlsing
-
3
"cost =
u =
i(z(s8
But in uni
,
this is written as z = lelcisO
s(z) 3
O
↑
(2) =
92 + 32
(2) =
5
tanf =
3
0 87.
=
36 .
z
=
g(cos(36 87) .
+ is in (36 87)) .
Solve for a :
x2 + 1 =
0
1
S
x2 =
x
=
=Fonti
↑
not real
Solve for a in :
d"-1 =
0
i) IR
x4 -
1 =
0
(x2+ 1)(x21) =
0 not iR
(x2 + 1)
(x + 1)(x 1)
-
=
0 ↑
x = = |[or = F]
ii)
x4 -
1 =
0
(x2+ 1)(x21) =
0 not iR
(x2 + 1)
(x + 1)(x 1)
-
=
0 ↑
x = = |[or = F]
iii) D (complex no . s
x4 -
1 =
0
(x2+ 1)(x[ 1) -
=
o
(x2 + 1)
1)(x 1) 0
=
(x +
-
x =
+
;t
iv) I
x4 -
1 =
0
(x2 1)(x[ 1)
+ -
=
o
(x2 + 1)
1)(x 1) 0
=
(x +
-
x =
1
Naming Complex Numbers
Symbols :
z
=
atbi
a + bi
1
real imaginary F
Ey : z
,
=
3 -
2)
z2
= 2+ gi
real part imaginery
↑ ↑
IRe (z , ) =
3 and Im(z ,) = -
2
IRe (zz) =
2 and
Im(zz) =
S
,Adding ↑ Subtracting Complex No S
.
1(z , zz + =
3 -
21 + 2 + 3i
=
5 + 3)
2 z
,
-
zz
=
(3 2i) (2 - si)
-
-
=
3 -
2i -
2 +
si
=
1 -
7
multiplying Complex No S .
3)z , xzy (3 2i)(z + gi) = -
4i -10 ;
2
=
b+ 1Si -
=
6 + 11) -
10) 1) -
=
6 + 11i + 10
=
10 + 11
in
Conjugate of z
If z = atbi
,
the conjugate,* =
a -
bi
Dividing Complex No S .
5) Z 3-2i
six2-si
=
~
↑ I
2+ 2 31
-
(3
-
-
zi)(z - si)
(2+ si) (z-si)
2
6-1Si-ui + ioi
?
4- loi + 10i 2si -
I
al
: Complex numbers are closed (ie : stay in the complex set) under +; -jx ; +
Argand Planes
x = Real
y Imaginary
=
Let z = 4 + 3)
im ·
(4 + 3i)
O
121
i
!
Y Re
U
length of z
modulus of z
= 121Y Argument of z : f
(a)
3
1 : =3 = Es tans-b 0 :
tan"
(i)
b
(z) =
5 cost =
1z) 0 =
cos"
| = an + b
=
sind i O sin"
:
(i)
, Modulus-Argument Form
z =
4 + 3i
z = (z) (cosf ising)+
-sinf =
ilzlsing
-
3
"cost =
u =
i(z(s8
But in uni
,
this is written as z = lelcisO
s(z) 3
O
↑
(2) =
92 + 32
(2) =
5
tanf =
3
0 87.
=
36 .
z
=
g(cos(36 87) .
+ is in (36 87)) .
