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Transcript
Complex Numbers
Modulus-Argument Form
Modulus-Argument Form
x  r cos
y  r sin 
Here is the real and imaginary part of the
The complex number z is marked on the
Argand diagram.
Im
z  x  yj
r
y

x
Re
complex number.
z

(
r
cos

)

(
r
sin

)
j
Cartesian coordinates are not the only
way to specify a position on a plane.
z  r (cos   j sin  )
The angle (argument) and distance from
This
is called
the modulus-argument
form
the origin
(modulus)
could also be used.
of the complex number.
Using simple trigonometry allows us to
find x and y in terms of r and θ.
Multiplication
u 2 32j
Im
v  1 j
uv
uv  (2 3  2 j )(1  j )
 2 32 3j2j2
v
u
 2( 3  1)  2( 3  1) j
Re
Argand diagram – the argument
Im
uv
v
u
Re
The modulus
u 2 32j
u  (2 3 ) 2  22  16  4
v  1 j
v  12  12  2
uv  2( 3  1)  2( 3  1) j
uv  (2( 3 1)) 2  (2( 3  1)) 2
 4(3  2 3  1)  4(3  2 3  1)
 32  4 2
The argument
arg(u )  arctan( 2 2 3 )


6
arg(v)  arctan( 11 )


5
 
6 4 12

4
arg(uv )  arctan( 2(
5

12
3 1)
2 ( 3 1)
)

Summary
When two complex numbers u and v are multiplied together, the modulus
of the product uv is equal to the modulus of u multiplied by the modulus of
v. The argument of uv is equal to the sum of the arguments of u and v.
uv  u v
arg( uv)  arg( u )  arg( v)
Division - modulus
u 2 32j
v  1 j
u 2 32j

v
1 j
2 3  2 j 1 j


1 j
1 j

2(1  3 )  2(1  3 ) j
2
 (1  3 )  (1  3 ) j
u
 (1  3 ) 2  (1  3 ) 2
v
 1 2 3  3 1 2 3  3
 82 2
u
4

2 2
v
2
Division - argument

arg(u ) 
6

arg(v) 
4
arg( u v )  arctan( 1

3

1 3
)
12
arg( u )  arg( v) 

6


4


12
Summary
When one complex number u is divided by another v, the modulus of u/v
is equal to the modulus of u divided by the modulus of v. The argument of
u/v is equal to the arguments of u minus the argument of v.
u u

v
v
arg( u v )  arg( u )  arg( v)
Using modulus-argument form
If complex numbers are written in modulus-argument form, it is easy to
find the modulus and argument of any product or quotient of the numbers
and hence the actual product and quotient.
u  3(cos 3  j sin 3 )
for
uv
u 3
v  2(cos 4  j sin 4 )
v 2
uv  3  2  6
arg( u)  3 arg( v)  4
hence
uv  6(cos 712  j sin
7
12
)
arg( uv)  3  4  712
Using modulus-argument form
w  5(cos 6  j sin
for
w
z
hence
z  cos 712  j sin
)

6
arg( z ) 
w
 5(cos 34  j sin
z
3
4
7
12
w
 5 1  5
z
z 1
w 5
arg( w) 

6
7
12
)
arg( wz ) 

6
 712 
9
12

3
4