Download de moivre`s theorem: powers and roots

Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
DE MOIVRE’S THEOREM:
POWERS AND ROOTS
ALGEBRA 8
INU0114/514 (MATHS 1)
Dr Adrian Jannetta MIMA CMath FRAS
De Moivre’s Theorem: powers and roots
1 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Objectives
This presentation will cover the following:
• Recap of polar and exponential form
• Understand more clearly the argument of a complex number
• De Moivre’s theorem
• Using De Moivre’s theorem to find powers and roots
De Moivre’s Theorem: powers and roots
2 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Recap: polar and Cartesian form
Argand diagrams allow a complex number z to be represented
geometrically.
The polar form is defined two
numbers.
yi
z = x + yi
The modulus (|z| or r) is the
distance r of the point from the
origin.
r
The argument (arg z or θ ) is an
θ
angle measured from the positive
x-axis.
We can convert from polar to Cartesian form using
x = r cos θ
y
x
x
and y = r sin θ
And from Cartesian form to polar form using
y
p
r = x2 + y 2 and θ = tan−1
x
where the argument θ must be in the correct quadrant.
De Moivre’s Theorem: powers and roots
3 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
A nice pattern...
Consider a complex number z in polar form:
z = cos θ + i sinθ
2
Let’s calculate z by squaring both sides:
z2 = (cos θ + i sinθ )2
Expand the brackets on the RHS:
z2
=
cos2 θ + i2 sin2 θ + 2i sinθ cos θ
=
cos2 θ − sin2 θ + 2i sinθ cos θ
We know from trigonometry that cos2 θ − sin2 θ ≡ cos 2θ and
2 sin θ cos θ ≡ sin 2θ .
Therefore:
z2 = cos 2θ + i sin2θ
To summarise:
(cos θ + i sinθ )2 ≡ cos 2θ + i sin2θ
We can take the power and multiply it with the angle. But is this true for
any power/exponent?
De Moivre’s Theorem: powers and roots
4 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
De Moivre’s Theorem
De Moivre’s theorem is a relationship
between complex numbers and
trigonometry.
Here it is:
(cos θ + i sin θ )n = cos nθ + i sin nθ
By expanding the LHS and comparing real
and imaginary coefficients it is possible to
derive trig identities for powers of sine and
cosine in terms of compound angles —
very useful in calculus.
Abraham de Moivre (1667 —
1754)
De Moivre’s Theorem: powers and roots
De Moivre’s theorem also makes it possible
to quickly evaluate powers and nth roots of
complex numbers.
5 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Evaluating powers
Calculating powers
Given the complex number
p
z = 1 + 3i
Use De Moivre’s theorem to evaluate z5 .
The modulus is |z| = 2 and arg z = π3 .
The polar form of z is
z = 2(cos π3 + i sin π3 )
Therefore:
z5
=
=
5
2(cos π3 + i sin π3 )
32(cos π3 + i sin π3 )5
By De Moivre’s theorem:
5π
z5 = 32(cos 5π
3 + i sin 3 )
Simplify to get the Cartesian form:
z5
=
=
De Moivre’s Theorem: powers and roots
p
3
2 i)
p
16 − 16 3i
32( 12 −
6 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
The argument revisited
yi
yi
yi
z = x + yi
r
θ
z = x + yi
r
r
θ + 2π
x
z = x + yi
θ + 2nπ
x
x
The argument of a complex number arg z = θ is the angle made by the
line to positive x-axis (first picture).
The angle θ can be increased by 2π and it will still describe the same line
in the xy-plane (second picture).
Any integer multiple of 2π can be added (or subtracted) to θ without
changing the Cartesian form of the complex number.
These statements are simply a consequence of the periodic nature of
sine and cosine:
cos(θ + 2nπ) ≡ cos θ
De Moivre’s Theorem: powers and roots
and sin(θ + 2nπ) ≡ sin θ
7 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Consider the complex number z = 4(cos π4 + i sin π4 ) = 4∠ π4 .
Here is the complex number on an Argand diagram
yi
z = 4(cos π4 + i sin π4 )
4
π
4
x
If we add 2π (or 4π or 6π or...2nπ) to the angle the point on the Argand
diagram will remain unchanged.
In this case z is also equivalent to
z = 4∠ 9π
4
The angle
π
4
, z = 4∠ 17π
4
, z = 4∠ 25π
4
is the principal value.
But adding multiples of 2π will give an equivalent angle.
De Moivre’s Theorem: powers and roots
8 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
The general polar form of a complex number is
z = r(cos(θ + 2πn) + i sin(θ + 2πn)) = r∠(θ + 2πn)
and the general exponential form is
z = ei(θ +2πn)
We have been restricting the allowable values to 0 ≤ θ < 2π. But as we
saw earlier, other values of θ are also valid.
Multivalued functions?
Previously we saw how to calculate logarithms of complex numbers. For
example
ln(−10) = ln 10 + πi
Recall that the definition of a function states that one input (in the
domain) should give one output value (in the range). We usually don’t
allow other values — like ln10 + 3πi — in order to satisfy that definition.
We do however, need to consider those other values of θ to obtain nth
roots using De Moivre’s theorem.
De Moivre’s Theorem: powers and roots
9 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Calculation of nth roots
Definition
A number w is a nth root of a number z if wn = z.
A few statements to clarify this definition...
• The number 2 is the 4th root of 16 because 24 = 16.
• The number −2 is also the 4th root of 16 because (−2)4 = 16.
• The number 3 is the cube root of 27 because 33 = 27.
From the previous example:
p
p
• The number 1 + 3i is the 5th root of 16 − 16 3i
The fundamental theorem of algebra tells to expect n roots — not just one. De
Moivre’s theorem in this case is
‹

‹‹


1
θ + 2πn
1
θ + 2πn
z n = r n cos
+ i sin
n
n
where n is an integer.
De Moivre’s Theorem: powers and roots
10 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Cube roots of a complex number
Find all of the cube roots of 27i.
This means we are looking for complex numbers z with the
property
z = 27i
First, we write z in polar form
z = 27(cos π2 + i sin π2 )
Take cube roots and apply De Moivre’s theorem:
1
z3
1
27(cos π2 + i sin π2 ) 3
=
1
= 3(cos π2 + i sin π2 ) 3
1
z3
= 3(cos π6 + i sin π6 )
1
In Cartesian form this is z 3 =
p
3 3
2
+ 32 i.
However, there are other values of z which give z = 27i.
De Moivre’s Theorem: powers and roots
11 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
We wrote 27i in polar form as
27i = 27(cos π2 + i sin π2 )
but we could also add 2π to the angle and it would still be 27i:
5π
27i = 27(cos 5π
2 + i sin 2 )
Therefore we have
5π
27i = z = 27(cos 5π
2 + i sin 2 )
If we apply De Moivre’s theorem to this:
1
z3
1
z3
=
1
5π 3
27(cos 5π
+
i
sin
)
2
2
5π
= 3(cos 5π
6 + i sin 6 )
1
p
In Cartesian form this is z 3 = − 3 2 3 + 23 i.
We found another cube root — different to the first!
De Moivre’s Theorem: powers and roots
12 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Add another 2π to the polar form for 27i and we get
z
1
z3
1
∴z3
=
9π
27(cos 9π
2 + i sin 2 )
=
9π 3
27(cos 9π
2 + i sin 2 )
=
3π
3(cos 3π
2 + i sin 2 )
1
1
In Cartesian form this is z 3 = −3i.
We’ve found three cube roots now. Are there any more?
Adding another 2π gives us
z
1
∴ z3
1
z3
=
13π
27(cos 13π
2 + i sin 2 )
=
13π
3(cos 13π
6 + i sin 6 )
=
p
3 3
2
+ 23 i
This is a repeat of the first root. If we add more multiples of 2π we will
also duplicate the other roots.
De Moivre’s Theorem: powers and roots
13 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
The number 27i has three cube roots. They are
z1 =
p
3 3
2
+ 23 i
p
z2 = − 3 2 3 + 23 i
z3 = −3i
Or in polar form:
z1 = 3∠ π6 ,
z2 = 3∠ 5π
6 ,
z3 = 3∠ 3π
2
On an Argand diagram we can see an obvious pattern with these roots:
yi
z2
z1
x
z3
The roots are equally spaced around the circle.
The angle between the roots on the diagram is
circle is |z| = 3.
De Moivre’s Theorem: powers and roots
2π
3 .
14 / 17
The radius of the
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
De Moivre’s Theorem and nth roots.
Calculate all of the 4th roots of −64
1
We are trying to find z 4 where z = −64. There will be four roots.
In polar form we write −64 = 64(cos π + i sin π) = 64∠π.
Therefore the four equations we must solve are:
z
=
64∠π
z
=
64∠3π
z
=
64∠5π
z
=
64∠7π
Applying De Moivre’s theorem to these gives the four roots:
p
1
1
z 4 = [64∠π] 4 = 8∠ π4 = 2 + 2i
p
1
1
z 4 = [64∠3π] 4 = 8∠ 3π
4 = −2 + 2i
p 5π
1
1
z 4 = [64∠5π] 4 = 8∠ 4 = −2 − 2i
p
1
1
z 4 = [64∠7π] 4 = 8∠ 7π
4 = 2 − 2i
De Moivre’s Theorem: powers and roots
15 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
The Argand diagram of the 4th roots of −64 looks like this:
yi
z2
z1
x
z3
z4
The roots are equally spaced around the circle; the angle between
p
π
them is 2π
4 = 2 . The radius of the circle is |z| = 8.
De Moivre’s Theorem: powers and roots
16 / 17
Adrian Jannetta
Introduction
De Moivre’s Theorem
Powers
Nth roots
Summary
Summary
Definition
A number w is a nth root of a number z if wn = z.
In general, to find the nth root of a complex number:
• There will be n roots
• The n roots will be equally spaced by an angle
2π
n .
• The polar form will be


‹

‹‹
1
1
θ + 2πn
θ + 2πn
z n = r n cos
+ i sin
n
n
where r is and θ is the principal value (of the first root).
De Moivre’s Theorem: powers and roots
17 / 17
Adrian Jannetta