Download Complex numbers in polar form

Complex numbers in polar
form
Modulus and argument
The point P( x, y ) can have its position fixed by
giving the distance OP and the angle OP makes with
the x-axis (in radians).
These quantities are respectively denoted by
r and  , and are called the polar coordinates of
the point P.
If we associate P with the complex number z in the Argand diagram,
then r is the same as the modulus of z i.e z .
The angle  is called the argument of z; it is denoted by arg z.
By convention, the argument is restricted so that       .
Furthermore, the argument is not defined if z  0.
Polar form
It is readily seen the x  r cos  and y  r sin  .
Hence, z  x  yi
 r cos    r sin   i
 r  cos   i sin   .
The non-zero complex number z can be written in modulus - argument form,
or polar form, as z  r  cos   i sin   , where r  z  0 is the modulus
and   arg z, with       , is the argument.
Examples
Do Q1, Q2, pp.242-243.
Write in modulus-argument form z  i, z  2, z  2  i and z  1  i.
z  i  r  1,  

2
z  2  r  2,   



 z  1 cos  i sin  .
2
2

 z  2  cos   i sin   .
z  2  i  r  5,   2.677  z  5  cos 2.677  i sin 2.677  .
z  1  i  r  2,   
3
4

 3
 z  2  cos  
 4


 3

i
sin



 4

 .

Example
If arg z 

4
and arg  z  3 

2
, find arg  z  6i  .
Since arg z 

4
, the point z lies on the half-line u
starting at O at an angle of
As arg  z  3 
angle of


2

4
to the real axis.
, the translation from 3 to z makes an
with the real axis, and so lies on the half-line v
2
in the direction of the imaginary axis.
The half-lines u and v meet at z  3  3i, so

arg  z  6i   arg(3  3i)   .
4
Do Q3-Q12, p.243.
Multiplication and division in polar
form
The rules for multiplication and division
in modulus-argument from are
st  s t ,
arg( st )  arg( s)  arg(t )  k (2 ),
s
arg    arg s  arg t  k (2 ),
t
where in each case the number k ( 1, 0,1)
is chosen to ensure that the argument lies
in the interval       .
s
s
 ,
t
t
Do Q1-Q8, p.247
Example
Let s   3  i and t  2  2i.
Then st  s t  2  2  4
5  13
and arg s  arg t 
 
.
6 4 12
11
11 

So st  4  cos
 i sin

12
12 

s 2
s
Also
  1
t
t 2
5  7
 
.
6 4 12
s
7
7
So  cos
 i sin
.
t
12
12
and arg s  arg t 
Do Q1, Q2, p.254
Square roots in polar
It is a consequence of the rule for multiplication that,
if s  p(cos   i sin  ), then s 2  p 2  cos 2  i sin 2  .
It is then easily seen that
s



p  cos  i sin  .
2
2

But we know that the two square roots are of the form  s .
So, the square roots of a complex number s
1
have modulus s and arguments arg s and
2
1
arg s   where the sign is positive if arg s  0
2
and negative if arg s  0.
Do Q1-Q11, pp.255-256