Download 5.7 Euler`s Marvelous Formula (slides, 4-to-1)

Euler’s Equation
The value of complex numbers was recognized but poorly understood
during the late Renaissance period (1500-1700 AD.) The number system
was explicitly studied in the late 18th century. Euler used i for the square
root of −1 in 1779. Gauss used the term “complex” in the early 1800’s.
Elementary Functions
Part 5, Advanced Trigonometry
Lecture 5.7a, Euler’s Marvelous Formula
The complex plane (“Argand diagram” or “Gauss plane”) was introduced
in a memoir by Argand in Paris in 1806, although it was implicit in the
doctoral dissertation of Gauss in 1799 and in work of Caspar Wessel
around the same time.
Dr. Ken W. Smith
Sam Houston State University
2013
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Euler’s Equation
Euler’s Equation
Notice the following remarkable fact that if
√
π
π
3 1
+ i = cos + i sin
z=
6
6
2
2
Euler would explain why that was true. Using the derivative and infinite
series, he would show that
then z 3 = i. (Multiply it out & see!) Thus z 12 = 1 and so z is a twelfth
root of 1.
Now the polar coordinate form for z is r = 1, θ = π6 , that is, z is exactly
one-twelfth of the way around the unit circle. z is a twelfth root of 1 and it
is one-twelfth of the way around the unit circle. This is not a coincidence!
DeMoivre apparently noticed this and proved (by induction, using sum of
angles formulas) that if n is an integer then
(cos θ + i sin θ)n = cos nθ + i sin nθ.
(1)
Thus exponentiation, that is raising a complex number to some power, is
equivalent to multiplication of the arguments. Somehow the angles in the
complex number act like exponents.
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eiθ = cos θ + i sin θ
(2)
By simple laws of exponents, (eiz )n = einz and so Euler’s equation
explains DeMoivre formula.
This explains the “coincidence” we noticed with the complex number
z = cos π6 + i sin π6 which is one-twelfth of the way around the unit circle;
raising z to the twelfth power will simply multiply the angle θ by twelve
and move the point z to the point with angle 2π: (1, 0) = 1 + 0i.
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Trig functions in terms of the exponential function
Trig functions in terms of the exponential function
Euler’s formula
We wrote the exponential function in terms of cosine and sine
eiθ = cos θ + i sin θ
allows us to write the exponential function in terms of the two basic trig
functions, sine and cosine. We may then use Euler’s formula to find a
formula for cos z and sin z as a sum of exponential functions.
By Euler’s formula, with input −z,
eiθ = cos θ + i sin θ
and then wrote the trig functions in terms of the exponential function!
e−iz = cos(−z) + i sin(−z) = cos(z) − i sin(z).
cos z =
eiz + e−iz
2
sin z =
eiz − e−iz
2i
Add the expressions for eiz and e−iz to get
eiz + e−iz = 2 cos(z)
and so
cos z =
eiz + e−iz
.
2
(3)
e−iz
If we subtract the equation
= cos z − i sin z from Euler’s equation
and then divide by 2i, we have a formula for sine:
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eiz
e−iz
−
sinElementary
z = Functions .
2i
2013
The exponential and trig functions are very closely related. Trig functions
are, in some sense, really exponential functions in disguise!
And conversely, the exponential functions are trig functions!
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(4)
Some worked examples.
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Some worked examples.
Let’s try out some applications of Euler’s formula. Here are some worked
problems.
Put the complex number z = eπi in the “Cartesian” form z = a + bi.
13π
i
6
Solution. z = eπi = 1(cos(π) + i sin(π)) = 1(−1 + 0i) = −1
Put the complex number z = 2e
It seems remarkable that if we combine the three strangest math
constants, e, i and π we get
Solution.
√
13π
13π
π
π
z = 2e 6 i = 2 cos( 13π
)
+
2i
sin(
)
=
2
cos(
)
+
2i
sin(
)
=
3 + i.
6
6
6
6
in the “Cartesian” form z = a + bi.
eπi = −1.
Some rewrite this in the form
eπi + 1 = 0
(often seen on t-shirts for engineering clubs or math clubs.)
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Some worked examples.
Some worked examples.
Put the complex number z = 18 + 26i in the “polar” form z = reiθ where
r, θ ∈ R and both r and θ are positive.
Solution. The modulus of z = 18 + 26i is 182 +√262 = 1000.
So the polar coordinate form of z = 18 + 26i is 103 eiθ where
26
). (The angle θ is about 0.96525166319.)
θ = arctan( 18
Find a cube root of the number z = 18 + 26i and put this cube root in the
“Cartesian” form z = a + bi. (Use a calculator and get an exact value for
this cube root.
√
Using the previous problem, we write z = 18 + 26i = 103 eiθ where
θ = arctan( 26
18 ). √
The cube root of 103 eiθ is
√
θ
10 ei 3
(The angle 3θ is about 0.3217505544.) Using a calculator, we can see that
this comes out to approximately
√
√
θ
θ
10 cos( ) + i 10 sin( ) = 3 + i.
3
3
One could check by computing (3 + i)3 and see that we indeed get
18 + 26i.
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Some worked examples.
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Some worked examples.
A question found on the internet: What is ii ?
π
We can find one answer if we write the base i in polar form i = e 2 i .
Find a complex number z such that ln(−1) = z.
π
Solutions. Since −1 in polar coordinate form is −1 = eiπ then z = πi is a
solution to ln(−1).
(More carefully, we might note that i = e 2 i+2πki , for any integer k.)
π
π 2
π
Then ii = (e 2 i )i = e 2 i = e− 2
≈
0.207879576350761908546955619834978770033877841631769608075135...
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Complex numbers v. Real numbers
Last Slide!
Here are some things one can do with the real numbers:
1
Show that f (x) = sin x is periodic with period 2π, that is,
f (x + 2π) = f (x).
2
Find an infinite set of numbers, x, such that sin(x) = 1/2.
3
Find a number x such that ex = 200.
4
Compute ln(2).
It is appropriate that we end our series of precalculus lectures with a
presentation of Euler’s marvelous formula, which brings together both the
trigonometric functions and the exponential functions into one form!
The applications of this formula appear in all the technology around us,
and simplify many complicated mathematical computations!
Here are some things that require complex numbers:
1
Show that f (x) = ex is periodic with period 2πi, that is,
f (x + 2πi) = f (x).
2
Find an infinite set of numbers, x, such that ex = 1/2.
3
Find a number x such that sin(x) = 200.
4
Compute ln(−2).
eiθ = cos θ + i sin θ
(End)
These are all topics for further exploration in a course in complex variables.
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