Download Unit 3.2 - Polar form and de Moivre`s Theorem The modulus of a

Unit 3.2 - Polar form and de Moivre’s Theorem
The modulus of a complex number plays an important role (Notes, p 61).
This concept is defined in such a way that in case the complex number happens to be real, the
modulus of this number turns out to be the absolute value of the number.
In the real case we know that the modulus of a number is equal to the distance of the number from
the point 0 on the real line.
In case of a complex number the modulus of the number turns out to be the distance between the
point representing this number in the Argand plane and the origin (the point (0,0)).
In this sense our new definition of modulus is just an extension of the same idea in the field of real
numbers.
You must know the properties of the modulus (Notes, Theorem 2, p 67).
We can use the modulus to sketch sets of complex numbers.
Example 1
Sketch all complex numbers z such that
|z + 1 − 2i| > 3.
Use Theorem 3 on p 67 to describe the modulus as the distance between two complex numbers:
|z + 1 − 2i| = |z − −1 + 2i |.
This is the distance between the complex number z and the complex number −1 + 2i.
You have to sketch all complex numbers z such that
|z − −1 + 2i | > 3
that is, such that the distance between z and −1 + 2i is greater than 3.
Using the Corollary on p 68 it follows that is the region outside of a circle with radius 3 and origin
in the complex plane at −1 + 2i.
00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
000000000000000000000000000000000000000000000000000
The next new concept is that of an argument of a complex number. Note that a number can have
many arguments, but only one main argument (Notes, p 69).
Consider the complex number z = a + ib. The number r is the modulus of the complex number
r = a 2 + b 2 and is the length of the line from the origin to the point.
The angle θ (sketch p 69) can be found by solving the two equations
cos θ = ar
and
sin θ = br .
It then follows that z = a + ib = rcos θ + i sin θ. This is the so-called polar form (Notes, p 69).
Be sure to be able to write any complex number in so-called polar form.
We will use the notation
rcos θ + i sin θ = rcisθ
for the polar form.
When you write complex numbers in polar form, always start with a sketch. Many times you can
find the value of θ from the sketch without solving the two equations.
Example 2
Write z = 2 − 2i in polar form.
2
2
-2
Use the formula to find r : r = |z| = 2 2 + −2  2 = 8 = 2 2 .
It follows from the sketch that the angle is − π (remember, you MUST use RADIANS for the
4
angle).
Therefore
z = 2 − 2i = 2 2 cos− π  + i sin− π  = 2 2 cis − π .
4
4
4
Example 2 on p 69 shows you how to use the formulas.
You can use polar form for multiplication (Notes, p 70) and division (Notes, p 72, Exercise 6), as
well as exponentiation of complex numbers (De Moivre’s Theorem, Notes, p 70).
As a matter of fact, multiplication boils down to addition, division of complex numbers actually
simplifies to subtraction, and exponentiation results into a simple multiplication.
Remember, however, that your first step is always to write the given complex number in polar
form.
Example 3
Write the complex number 2 − 2i 4 in the form a + ib.
From the sketch above, the polar form of z = 2 − 2i is z = 2 2 cis − π .
4
Therefore
z 4 = [2 2 cis − π  4
4
4
= [2 2  cis 4 × − π 
4
= 64 cis −π
= 64cos−π + i sin−π
= −64.
An important consequence of de Moivre’s Theorem, is the possibility of calculating the nth roots
of any real or complex number (Notes, Theorem 5, p 72).
It also turns out that every number has exactly n different nth roots.
I am giving you another wording for the theorem:
Let n be a natural number and consider the nonzero complex number, z,
z = rcos θ + i sin θ = r cis θ.
The equation
xn = z
has exactly n different solutions, namely
x = wk =
n
r cis  θ +n2kπ 
where k = 0, 1, ... ,n − 1.
You should make sure that you can write down the sine or cosine of angles such as 0,
π, π , π , π , and π . You should also be sure of the correct sign of these expressions in all
2 3 4
6
the different quadrants.
When applying de Moivre’s Theorem one has to write down n r , where r is the modulus of the
complex number. Do remember that your answer should always be a positive real number. Thus
the expression −2(cos π + i sin π is NOT in polar form.
Example 4
Solve x 2 = −4i in
C.
Use a sketch to write the complex number −4i in polar form.
x 2 = −4i = 4cis− π .
2
Use the formula above to find the roots:
x = wk =
n
r cis  θ +n2kπ  =
4 cis 
− π + 2kπ
2
2
.
Now simplify the formula:
− π + 2kπ
π + kπ, k = 0, 1.
x = wk = 4 cis  2
 = 2cis−
2
4
Substitute the values of k :
x = w0 = 2cis− π  = 2cos− π  + i sin− π  = 2 − i 2
4
4
4
3
π
π
π
3
3
x = w1 = 2cis
 = 2cos
+ i sin
 = − 2 +i 2
4
4
4
Example 5
Find the 3-rd roots of −4 3 − 4i.
If z = −4 3 − 4i then r = 64 = 8 and
cos θ =
−4 3
8
=
− 3
sin θ = −4 = −1
8
2
2

θ = π + π = 7π
6
6
The MAIN ARGUMENT is θ = 7π − 2π = − 5π and
6
6
5
π
5
π
5π .
−4 3 − 4i = 8cos−
 + i sin−
 = 8 cis−
6
6
6
The third roots of −4 3 − 4i are the solutions of the equation z 3 = 8 cis− 5π .
6
r cis  θ +n2kπ 
− 5π + 2kπ
3

= 8 cis  6
3
= 2 cis − 5π + 2kπ
you MUST simplify the argument
18
3
= 2 cis −5π + 12kπ
, k = 0, 1, 2.
18
x = wk =
n
x = w0 = 2 cis
− 5π
or x = w1 = 2 cis
18
31
π
or x = w3 = 2 cis
.
18
7π
18
or x = w2 = 2 cis
19π
18
Lastly you should have a close look at the geometric representation of the nth roots of a complex
number on a circle.
Do example 5 on p 72.
You should now be ready for the last theme, where we will factorize polynomials in different ways.
Where previously we were satisfied to say that the equation x 4 + 1 = 0 has no roots, we now can
write down four different roots - of course they will all be complex numbers. Can you write down
these roots straight away?