ω is the Greek letter omega.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
You do not need to be able to prove
theorems but you will need to use the
concepts in order to understand this
module.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
Abraham Demoivre (1667–1754) French
mathematician and demographer. (The
name is sometimes written as de Moivre.)
Being Protestant, he fled to England on the
revocation of the Edict of Nantes, and
supported himself by teaching. He was
chosen to be the judge in the famous
contest between Newton and Leibniz for
the credit of the invention of calculus.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
To see that
i − 1 = 21/2[cos1(3π/4) + i1sin1(3π/4)]
plot i − 1 on an Argand diagram, and note
that
|1i − 11| = 21/2 while arg1(i − 1) = 3π/4
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
There is no need to check this calculation;
it is given as an example of how not to
calculate z8 .
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
The binomial theorem provides an easier
way of expanding expressions such as
z +

1
z
3
(see Subsection 3.4).
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
This expansion of (z – 1/z)7 could be
obtained by multiplying the expression out
at length, although it would be much
simpler to use the binomial theorem
(discussed in Subsection 3.4). Here you
can take the expansion on trust.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
If you are wondering why it must be an
odd power of sin1θ then try the method on
sin 4 1 θ and you will soon discover the
reason.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
Again the expansion can be obtained from
the binomial theorem.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
i02 = −1 so that i04 = 1.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
Multiplying
3e2i × 5e7 0 i = 15e 9i
Squaring
5e7i × 5e7i = 25e14i
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
See Question R2.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
We must use the general expression e2πki
in order to find all the roots of the
equation. If we use just one value of k then
we will obtain only one root, as for
example when we found just one of the
square roots of 1 + i in Question T2.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
See Question R2.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
γ is the Greek letter gamma.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
ε is the Greek letter epsilon.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
For a.c. circuits, R is the resistance and
X is the reactance.
Don’t worry if you are unfamiliar with
these terms; you will come across them
later in your physics courses.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
In spite of the simplicity of this result it
is relevant to the physical example of an
a.c. circuit in which impedances connected
in series.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
This is more of a test of your skill with
algebra than your understanding of
complex numbers!
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
The binomial expansion is also known as
the binomial theorem or the binomial
series.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
If you are not familiar with a proof of the
binomial theorem, see the Glossary for
details.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
The proof of this result is identical to that
for the real case.
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
Multiplying the numerator and
denominator by 1 − e−iθ (the conjugate
of 1 + eiθ ) then writing
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
(1 + eiθ )(1 − e−iθ) = 2(1 − cos1θ)
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
i – 1 = 21/2[cos1(3π/4) + i1sin1(3π/4)] ☞
and therefore, using Demoivre’s theorem
with n = 1/2,
(i − 1)1/2 =
2 1/4[cos1(3π/8) + i1sin1(3π/8)].4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
Let z = eiθ then
sin 5 θ
=
1 
1
z− 
z
(2i) 5 
5
1 
 5 1 
z − 5 − 5 z 3 − 3  

z 
z 
i 
=−

32 
1


 +10  z − 
z


i  2isin (5θ ) − 5 × 2i sin (3θ ) 
=− 

32  +10 × 2i sin θ

☞
=
1
[sin (5θ ) − 5sin (3θ ) + 10sin θ ] 4❏
16
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
From Demoivre’s theorem, with n = 5,
we have
cos1(5θ ) + i1sin1(5θ) = (cos1 θ + i1sin1θ)5
= cos51 θ + 5i1cos4 1 θ1sin1θ – 101cos3 1 θ1sin21 θ–
10i1cos2 1 θ sin3 1 θ + 5cos1 θ 1sin4 1 θ + i1sin51 θ
☞
and equating real and imaginary parts
gives us
cos1(5θ ) = cos51 θ – 101cos3 1 θ 1sin21 θ
+ 51cos1 θ 1sin4 1 θ
and
sin1(5θ) = 51cos4 1 θ1sin1θ
– 101cos2 1 θ sin3 1 θ + sin5 1 θ4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
First we write the complex number in
exponential form 1 + i = 2 e i π 4 , then we
take the cube root of (the real number) 2
and divide π/4 by three to give
(1 + i)1 3 = 21 6 e i π 12 .
☞4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
Writing 1 + i in the alternative exponential
form we have 1 + i = 2 e i π 4 e 2π k i
where k = 0, 1, 2. The three values are
z 0 = 21 6 e i π 12 e 0i = 21 6 e i π 12 ,
z1 = 21 6 e i π 12 e 2i π 3 = 21 6 e 3i π 4 ,
and
z2 = 21 6 e i π 12 e 4i π 3 = 21 6 e17i π 12 .
☞4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
In this case (and using the usual notation
for a quadratic equation)
a = 1, b = 1 and c = i.
The roots of the equation are therefore,
z=
−b ± b 2 − 4ac −1 ± 1 − 4i
=
2a
2
However, we can find a square root of
1 – 4i by writing
1 − 4i = 4.1231e−1.3258i
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
so that
1 − 4i ≈ (4.1231)1 2 e −0.6629i
≈ 2.0305[cos(0.6629) − isin (0.6629)]
≈ 1.6005 − 1.2496i
and therefore the roots of the equation are
−1 ± (1.6005 − 1.2496i)
which gives
2
0.300 – 0.625i4and4–1.300 + 0.625i
as the roots of the equation (which we can
easily check by substituting these values
into the original equation).4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
Multiplying the first equation by i and the
second by 2 we obtain
2i0z − w = 5i − 1
(23)
2i0z − 6w = 2i
(24)
Subtracting Equation 24 from Equation 23
we obtain
5w = 3i – 1 so that w =
3i − 1
5
and substituting this value for w into
Equation 22 we have
z = −3iw + 1 = 1 − 3i 

3i − 1  3i + 14
=
4❏
5
5 
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
3 + 2i
1
−
1 + i 1 + 2i
3 + 2i 1 − i
1
1 − 2i
=
×
−
×
1 + i 1 − i 1 + 2i 1 − 2i
=
5 − i 1 − 2i 23 − i
−
=
4❏
2
5
10
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
3
r
3 
1 
 z − 1 =
3
C3−r z 3−r ×  −  

∑

 z 
z
r=0 


z2 3
z
1
+ C1 2 − 3C0 3
z
z
z
3 1
= z 3 − 3z + − 3
z z
= 3C3 z 3 − 3C2
1
1
=  z 3 − 3  − 3 z −  4❏

z
z  
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
7
r
7 
1 
 z − 1 =
7
C7−r z 7−r ×  −  

∑

 z 
z

r=0 

= z 7 − 7z 5 + 21z 3 − 35z +
35 21 7
1
−
+
−
z z3 z5 z7
1
1
1
=  z 7 − 7  − 7 z 5 − 5  + 21 z 3 − 3 



z 
z 
z 
1
− 35 z − 

z
4❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
1 + e i θ + e 2i θ + e 3i θ =
1 − e 4i θ
4❏
1 − eiθ
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
✧
Let z = eiθ then
sin1(θ 0) + sin1(2θ0) + sin1(3θ0) + … + sin1(9θ0)
= Im(z + z2 + … + z 9 )
 e iθ − e10iθ 
 z − z10 
= Im 
 = Im 

 1− z 
 1 − e iθ 
 (e iθ − e10iθ )(1 − e −iθ ) 
= Im 

iθ
−iθ
 (1 − e )(1 − e ) 
☞
1
Im(e iθ − 1 − e10iθ + e 9iθ )
2(1 − cos θ )
sin θ − sin (10 θ ) + sin (9θ )
4
=
2(1 − cos θ )
=
❏
FLAP M3.3
Demoivre’s theorem and complex algebra
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1