Download Complex Numbers - Jonas Oppenheim

Calculus Notes:
Complex Numbers
Jonas Oppenheim†
MATH 022 - Calculus II
August 8, 2016
1
Standard Form
Standard Form:
z = a + bi
√
where a, b ∈ R and i = −1. If a = 0, then the complex number is pure imaginary. If b = 0,
then the complex number is real. Note that R ⊆ C.
If z = a + bi and a, b ∈ R, then the conjugate of z is
z̄ = a − bi
and, as is seen figure 1, the modulus (or absolute value) of z is
√
|z| = a2 + b2
1.1
Arithmetic
Let
w = −5 + 6i
z = 3 − 8i.
1.1.1
Addition
Simply add the real and imaginary terms independently:
w + z = (−5 + 3) + (6i − 8i)
= −2 − 2i.
†
Jonas N. Oppenheim, Calculus II, University of Vermont, Burlington, VT 05405; [email protected];
http://www.jonasoppenheim.com
1
Figure 1: The modulus of z is |z|.
1.1.2
Subtraction
Subtract the real and imaginary terms independently:
w − z = (−5 − 3) + (6i + 8i)
= −8 + 14i.
1.1.3
Multiplication
Grit your teeth and distribute:
w × z = (−5 + 6i)(3 − 8i)
= −15 + 40i + 18i − 48i2
= 33 + 58i.
Note that since i =
1.1.4
√
−1, i2 = −1.
Division
Multiply the numerator and denominator by the conjugate of the denominator:
w
−5 + 6i
=
z
3 − 8i
(−5 + 6i)(3 + 8i)
=
(3 − 8i)(3 + 8i)
−15 − 40i + 18i + 48i2
=
9 + 24i − 24i − 64i2
−63 − 22i
=
73
−63 22i
=
−
73
73
Note that if z = a + bi, then z × z̄ = (a + bi)(a − bi) = |z|2 .
2
2
Polar Form
Polar form:
z = r(cos(θ) + i sin(θ))
where
r = |z|
θ = Arg(z).
Alternatively, Euler’s Formula shows that the polar form may also be given as
z = reiθ .
(a) T1 (x)
(b) T2 (x)
Figure 2: Comparing
3
Example Problem
Let
z = 5 − 5i.
a) Express z in polar form.
Begin by plotting z on the complex coordinate plane. Recall that polar form is z =
r(cos(θ) + i sin(θ)) and so we need to find r and θ. In order to find r, we use the Pythagorean
3
theorem. In this case, we find that
r 2 = a2 + b 2
r2 = (5)2 + (−5)2
√
r = 50
√
= 5 2.
Next, we find θ by recognizing the type
√ of triangle being formed on the complex coordinate
plane. In this case, we have a 1, 1, 2 triangle and so θ = 45 degrees = π4 . Note, though,
or −π
. Now that we know r and θ,
that because r is positive, θ must be given as either 7π
4
4
√
7π
7π
we know that the polar form of 5 − 5i is 5 2(cos( 4 + i sin( 4 )).
b) Find z 12 .
We will proceed using the Euler form of z, so that
√ 7π
z = 5 2ei 4 .
Raising both sides to the power of 12, we have
√ 7π
z 12 = (5 2ei 4 )12
= 512 26 ei21π
Now put the result back into polar form:
512 26 ei21π = 512 26 (cos(21π) + i sin(21π))
= 512 26 (−1 + i0)
= −(512 26 ).
4
Euler’s Formula
ex =
∞
X
xn
n=0
iθ
e =
∞
X
(iθ)n
n=0
n!
=1+
n!
iθ i2 θ2 i3 θ3 i4 θ4 i5 θ5 i6 θ6 i7 θ7
+
+
+
+
+
+
+ ...
1!
2!
3!
4!
5!
6!
7!
iθ θ2 iθ3 θ4 iθ5 θ6 iθ7
−
−
+
+
−
−
+ ...
1!
2!
3!
4!
5!
6!
7!
θ2 θ4 θ6
θ
θ3 θ5 θ7
= (1 −
+
−
+ . . . ) + i( −
+
−
+ ...)
2!
4!
6!
1! 3!
5!
7!
=1+
4
eiθ = cos(θ) + i sin(θ)
Now strap yourself in and consider the case where θ = π. Since cos(π) = −1 and sin(π) = 0,
the clouds part, we are bathed in sunlight, and a stone tablet descends from the heavens
that reads
eiπ = −1.
5