Modulus-argument form
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Polar form of a complex number
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It is possible to express complex numbers in polar form. The complex number z in the diagram below can be described by the length r and the angle
θ
{\displaystyle \theta }
[Argand diagram]
The distance r is the modulus of z,
|
z
|
{\displaystyle |z|\,}
θ
{\displaystyle \theta }
2
π
{\displaystyle 2\pi }
arg
z
{\displaystyle \arg {z}\,}
−
π
<
θ
≤
π
{\displaystyle -\pi <\theta \leq \pi }
The following Argand diagram shows the complex number
z
=
1
+
3
j
{\displaystyle z=1+{\sqrt {3}}j}
|
z
|
=
1
2
+
3
2
=
1
+
3
=
2
{\displaystyle |z|={\sqrt {1^{2}+{\sqrt {3}}^{2}}}={\sqrt {1+3}}=2}
arg
z
=
arctan
3
1
=
π
3
{\displaystyle \arg z=\arctan {\frac {\sqrt {3}}{1}}={\frac {\pi }{3}}\,}
This Argand diagram shows the complex number
z
=
2
−
4
j
{\displaystyle z=2-4j\,}
This Argand diagram shows the complex number
z
=
−
8
+
5
j
{\displaystyle z=-8+5j\,}
This Argand diagram shows the complex number
z
=
−
5
−
6
j
{\displaystyle z=-5-6j\,}
When we have a complex number
z
=
x
+
y
j
{\displaystyle z=x+yj\,}
(
r
,
θ
)
{\displaystyle (r,\theta )\,}
x
=
r
cos
θ
{\displaystyle x=r\cos {\theta }\,}
y
=
r
sin
θ
{\displaystyle y=r\sin {\theta }\,}
z
=
r
(
cos
θ
+
j
sin
θ
)
{\displaystyle z=r(\cos {\theta }+j\sin {\theta })\,}
Multiplication and division
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The polar form of complex numbers can provide a geometrical interpretation of the multiplication and division of complex numbers.
Multiplication
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Take two complex numbers in polar form,
ω
1
=
r
1
(
cos
θ
1
+
j
sin
θ
1
)
{\displaystyle \omega _{1}=r_{1}(\cos {\theta _{1}}+j\sin {\theta _{1}})\,}
ω
2
=
r
2
(
cos
θ
2
+
j
sin
θ
2
)
{\displaystyle \omega _{2}=r_{2}(\cos {\theta _{2}}+j\sin {\theta _{2}})\,}
and then multiply them together,
ω
1
ω
2
=
r
1
r
2
(
cos
θ
1
+
j
sin
θ
1
)
(
cos
θ
2
+
j
sin
θ
2
)
=
r
1
r
2
(
(
cos
θ
1
cos
θ
2
−
sin
θ
1
sin
θ
2
)
+
j
(
sin
θ
1
cos
θ
2
+
cos
θ
1
sin
θ
2
)
)
=
r
1
r
2
(
cos
(
θ
1
+
θ
2
)
+
j
sin
(
θ
1
+
θ
2
)
)
{\displaystyle {\begin{array}{rl}\omega _{1}\omega _{2}&=r_{1}r_{2}(\cos {\theta _{1}}+j\sin {\theta _{1}})(\cos {\theta _{2}}+j\sin {\theta _{2}})\\&=r_{1}r_{2}((\cos {\theta _{1}}\cos {\theta _{2}}-\sin {\theta _{1}}\sin {\theta _{2}})+j(\sin {\theta _{1}}\cos {\theta _{2}}+\cos {\theta _{1}}\sin {\theta _{2}}))\\&=r_{1}r_{2}(\cos {(\theta _{1}+\theta _{2})}+j\sin {(\theta _{1}+\theta _{2})})\end{array}}\,}
The result is a complex number with a modulus of
r
1
r
2
{\displaystyle r_{1}r_{2}\,}
θ
1
+
θ
2
{\displaystyle \theta _{1}+\theta _{2}\,}
|
ω
1
ω
2
|
=
|
ω
1
|
|
ω
2
|
{\displaystyle |\omega _{1}\omega _{2}|=|\omega _{1}||\omega _{2}|\,}
arg
(
ω
1
ω
2
)
=
arg
ω
1
+
arg
ω
2
{\displaystyle \arg {(\omega _{1}\omega _{2})}=\arg {\omega _{1}}+\arg {\omega _{2}}\,}
Division
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Dividing two complex number
z
1
{\displaystyle z_{1}\,}
z
2
{\displaystyle z_{2}\,}
z
1
=
r
1
(
c
o
s
θ
1
+
i
sin
θ
1
)
{\displaystyle z_{1}=r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})\,}
z
2
=
r
2
(
c
o
s
θ
2
+
i
sin
θ
2
)
{\displaystyle z_{2}=r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})\,}
⇒
z
1
z
2
=
r
1
(
c
o
s
θ
1
+
i
sin
θ
1
)
r
2
(
c
o
s
θ
2
+
i
sin
θ
2
)
{\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})}{r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})}}}
Multiply numerator and denominator by
(
c
o
s
θ
2
−
i
sin
θ
2
)
{\displaystyle (cos{\theta _{2}}-i\sin {\theta _{2}})}
=
r
1
(
c
o
s
θ
1
+
i
sin
θ
1
)
(
c
o
s
θ
2
−
i
sin
θ
2
)
r
2
(
c
o
s
θ
2
+
i
sin
θ
2
)
(
c
o
s
θ
2
−
i
sin
θ
2
)
{\displaystyle ={\frac {r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})(cos{\theta _{2}}-i\sin {\theta _{2}})}{r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})(cos{\theta _{2}}-i\sin {\theta _{2}})}}}
Then, use distribution to simplify.
=
r
1
(
c
o
s
θ
1
cos
θ
2
−
i
cos
θ
1
sin
θ
2
+
i
sin
θ
1
cos
θ
2
−
i
2
sin
θ
1
sin
θ
2
)
r
2
(
c
o
s
2
θ
2
−
i
sin
θ
2
cos
θ
2
+
i
sin
θ
2
cos
θ
2
−
i
2
sin
2
θ
2
)
{\displaystyle ={\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}-i\cos {\theta _{1}}\sin {\theta _{2}}+i\sin {\theta _{1}}\cos {\theta _{2}}-i^{2}\sin {\theta _{1}}\sin {\theta _{2}})}{r_{2}(cos^{2}{\theta _{2}}-i\sin {\theta _{2}}\cos {\theta _{2}}+i\sin {\theta _{2}}\cos {\theta _{2}}-i^{2}\sin ^{2}{\theta _{2}})}}}
Here, factorize by
i
{\displaystyle i}
i
2
=
−
1
{\displaystyle i^{2}=-1}
r
1
(
c
o
s
θ
1
cos
θ
2
−
i
2
sin
θ
1
sin
θ
2
+
i
(
s
i
n
θ
1
cos
θ
2
−
c
o
s
θ
1
sin
θ
2
)
r
2
(
c
o
s
2
θ
2
+
s
i
n
2
θ
2
)
{\displaystyle {\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}-i^{2}\sin {\theta _{1}}\sin {\theta _{2}}+i(sin{\theta _{1}}\cos {\theta _{2}}-cos{\theta _{1}}\sin {\theta _{2}})}{r_{2}(cos^{2}{\theta _{2}}+sin^{2}{\theta _{2}})}}}
=
r
1
(
c
o
s
θ
1
cos
θ
2
+
s
i
n
θ
1
sin
θ
2
+
i
(
s
i
n
θ
1
cos
θ
2
−
c
o
s
θ
1
sin
θ
2
)
r
2
(
1
)
{\displaystyle ={\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}+sin{\theta _{1}}\sin {\theta _{2}}+i(sin{\theta _{1}}\cos {\theta _{2}}-cos{\theta _{1}}\sin {\theta _{2}})}{r_{2}(1)}}}
Apply the formulas for the cosine of the difference of two angles and for the sine of the difference of two angles:
z
1
z
2
=
r
1
r
2
(
c
o
s
(
θ
1
−
θ
2
)
+
i
sin
(
θ
1
−
θ
2
)
)
{\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}}{r_{2}}}(cos{(\theta _{1}-\theta _{2})}+i\sin {(\theta _{1}-\theta _{2})})}
De Moivre's theorem
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Using the multiplication rules we can see that if
z
=
cos
θ
+
j
sin
θ
{\displaystyle z=\cos {\theta }+j\sin {\theta }\,}
then
z
2
=
cos
2
θ
+
j
sin
2
θ
{\displaystyle z^{2}=\cos {2\theta }+j\sin {2\theta }\,}
z
3
=
cos
3
θ
+
j
sin
3
θ
{\displaystyle z^{3}=\cos {3\theta }+j\sin {3\theta }\,}
De Moivre's theorem states that this holds true for any integer power. So,
z
n
=
cos
n
θ
+
j
sin
n
θ
{\displaystyle z^{n}=\cos {n\theta }+j\sin {n\theta }\,}
Complex exponents
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Definition
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If we let
z
=
cos
θ
+
j
sin
θ
{\displaystyle z=\cos {\theta }+j\sin {\theta }\,}
θ
{\displaystyle \theta }
d
z
d
θ
=
−
sin
θ
+
j
cos
θ
=
j
2
sin
θ
+
j
cos
θ
=
j
(
cos
θ
+
j
sin
θ
)
=
j
z
{\displaystyle {\begin{aligned}{\frac {dz}{d\theta }}&=-\sin {\theta }+j\cos {\theta }\\&=j^{2}\sin {\theta }+j\cos {\theta }\\&=j(\cos {\theta }+j\sin {\theta })\\&=jz\end{aligned}}}
The general solution to the differential equation
d
z
d
θ
=
j
z
{\displaystyle {\frac {dz}{d\theta }}=jz}
z
=
e
j
θ
+
c
{\displaystyle z=e^{j\theta +c}\,}
This means that
cos
θ
+
j
sin
θ
=
e
j
θ
+
c
{\displaystyle \cos {\theta }+j\sin {\theta }=e^{j\theta +c}\,}
By putting
θ
{\displaystyle \theta }
cos
0
+
j
sin
0
=
e
0
+
c
1
+
0
j
=
e
c
⇒
c
=
0
{\displaystyle {\begin{array}{rrcl}&\cos {0}+j\sin {0}&=&e^{0+c}\\&1+0j&=&e^{c}\\\Rightarrow &c&=&0\end{array}}}
So the general definition can be made:
e
j
θ
=
cos
θ
+
j
sin
θ
{\displaystyle e^{j\theta }=\cos {\theta }+j\sin {\theta }\,}
For a complex number
z
=
x
+
y
j
{\displaystyle z=x+yj\,}
e
z
{\displaystyle e^{z}\,}
e
z
=
e
x
+
y
j
=
e
x
e
y
j
=
e
x
(
cos
y
+
j
sin
y
)
{\displaystyle e^{z}=e^{x+yj}=e^{x}e^{yj}=e^{x}(\cos {y}+j\sin {y})\,}
Proof of de Moivre's theorem
edit
We can now give an alternative proof of de Moivre's theorem for any rational value of n:
(
cos
θ
+
j
sin
θ
)
n
=
(
e
j
θ
)
n
=
e
j
n
θ
=
e
j
(
n
θ
)
=
cos
n
θ
+
j
sin
n
θ
{\displaystyle {\begin{aligned}(\cos {\theta }+j\sin {\theta })^{n}&=(e^{j\theta })^{n}\\&=e^{jn\theta }\\&=e^{j(n\theta )}\\&=\cos {n\theta }+j\sin {n\theta }\\\end{aligned}}}
Summations
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deMoivre's therom can come in handy for finding simple expressions for infinite series. This usually involves a series multiple angles, cosrθ, as in this next example:
Infinite series are defined by:
C
=
c
o
s
2
θ
−
1
2
c
o
s
5
θ
+
1
4
c
o
s
8
θ
−
1
8
c
o
s
11
θ
+
.
.
.
{\displaystyle C=cos2\theta -{\frac {1}{2}}cos5\theta +{\frac {1}{4}}cos8\theta -{\frac {1}{8}}cos11\theta +...}
S
=
s
i
n
2
θ
−
1
2
s
i
n
5
θ
+
1
4
s
i
n
8
θ
−
1
8
s
i
n
11
θ
+
.
.
.
{\displaystyle S=sin2\theta -{\frac {1}{2}}sin5\theta +{\frac {1}{4}}sin8\theta -{\frac {1}{8}}sin11\theta +...}
In order to find either the sum of C or the sum of S (or both!) you need to add C to jS:
C
+
j
S
=
c
o
s
2
θ
+
j
s
i
n
2
θ
−
1
2
(
c
o
s
5
θ
+
j
s
i
n
5
θ
)
+
1
4
(
c
o
s
8
θ
+
j
s
i
n
8
θ
)
−
1
8
(
c
o
s
11
θ
+
j
s
i
n
11
θ
)
{\displaystyle C+jS=cos2\theta +jsin2\theta -{\frac {1}{2}}\left(cos5\theta +jsin5\theta \right)+{\frac {1}{4}}\left(cos8\theta +jsin8\theta \right)-{\frac {1}{8}}\left(cos11\theta +jsin11\theta \right)}
C
+
j
S
=
(
c
o
s
θ
+
j
s
i
n
θ
)
2
−
1
2
(
c
o
s
θ
+
j
s
i
n
θ
)
5
+
1
4
(
c
o
s
θ
+
j
s
i
n
θ
)
8
−
1
8
(
c
o
s
θ
+
j
s
i
n
θ
)
11
{\displaystyle C+jS=(cos\theta +jsin\theta )^{2}-{\frac {1}{2}}\left(cos\theta +jsin\theta \right)^{5}+{\frac {1}{4}}\left(cos\theta +jsin\theta \right)^{8}-{\frac {1}{8}}\left(cos\theta +jsin\theta \right)^{11}}
It is easier to work with now using the form e^jθ:
C
+
j
S
=
e
2
j
θ
−
1
2
e
5
j
θ
+
1
4
e
8
j
θ
+
.
.
.
{\displaystyle C+jS=e^{2j\theta }-{\frac {1}{2}}e^{5j\theta }+{\frac {1}{4}}e^{8j\theta }+...}
and you should be able to see the pattern, that the factor is negative 1/2 to the power of n-1 (the negative being alternately to even then odd powers is what makes it flip between + and -) and the power of e (the number in front of θ in our original equations) is equal to 3(n-1)jθ.
C
+
j
S
=
−
(
−
1
2
)
n
−
1
e
(
3
n
−
1
)
j
θ
=
(
−
1
2
e
3
j
θ
)
n
−
1
{\displaystyle C+jS=-\left({\frac {-1}{2}}\right)^{n-1}e^{(3n-1)j\theta }=\left({\frac {-1}{2}}e^{3j\theta }\right)^{n-1}}
This is a geometric series, with a=
Complex roots
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Applications of complex numbers in geometry
edit
