Elementary FunctionsPart 5, Advanced Trigonometry

Lecture 5.7a, Euler’s Marvelous Formula

Dr. Ken W. Smith

Sam Houston State University

2013

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Euler’s Equation

The value of complex numbers was recognized but poorly understoodduring the late Renaissance period (1500-1700 AD.) The number systemwas explicitly studied in the late 18th century. Euler used i for the squareroot of −1 in 1779. Gauss used the term “complex” in the early 1800’s.

The complex plane (“Argand diagram” or “Gauss plane”) was introducedin a memoir by Argand in Paris in 1806, although it was implicit in thedoctoral dissertation of Gauss in 1799 and in work of Caspar Wesselaround the same time.

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Euler’s Equation

Notice the following remarkable fact that if

z =

√3

2+

1

2i = cos

π

6+ i sin

π

6

then z3 = i. (Multiply it out & see!) Thus z12 = 1 and so z is a twelfthroot of 1.Now the polar coordinate form for z is r = 1, θ = π

6 , that is, z is exactlyone-twelfth of the way around the unit circle. z is a twelfth root of 1 and itis one-twelfth of the way around the unit circle. This is not a coincidence!DeMoivre apparently noticed this and proved (by induction, using sum ofangles formulas) that if n is an integer then

(cos θ + i sin θ)n = cosnθ + i sinnθ. (1)

Thus exponentiation, that is raising a complex number to some power, isequivalent to multiplication of the arguments. Somehow the angles in thecomplex number act like exponents.

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Euler’s Equation

Euler would explain why that was true. Using the derivative and infiniteseries, he would show that

eiθ = cos θ + i sin θ (2)

By simple laws of exponents, (eiz)n = einz and so Euler’s equationexplains DeMoivre formula.

This explains the “coincidence” we noticed with the complex numberz = cos π6 + i sin π

6 which is one-twelfth of the way around the unit circle;raising z to the twelfth power will simply multiply the angle θ by twelveand 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

Euler’s formulaeiθ = cos θ + i sin θ

allows us to write the exponential function in terms of the two basic trigfunctions, sine and cosine. We may then use Euler’s formula to find aformula for cos z and sin z as a sum of exponential functions.By Euler’s formula, with input −z,

e−iz = cos(−z) + i sin(−z) = cos(z)− i sin(z).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)

If we subtract the equation e−iz = cos z − i sin z from Euler’s equationand then divide by 2i, we have a formula for sine:

sin z =eiz − e−iz

2i. (4)Smith (SHSU) Elementary Functions 2013 5 / 14

Trig functions in terms of the exponential function

We wrote the exponential function in terms of cosine and sine

eiθ = cos θ + i sin θ

and then wrote the trig functions in terms of the exponential function!

cos z =eiz + e−iz

2

sin z =eiz − e−iz

2i

The exponential and trig functions are very closely related. Trig functionsare, in some sense, really exponential functions in disguise!

And conversely, the exponential functions are trig functions!

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Some worked examples.

Let’s try out some applications of Euler’s formula. Here are some workedproblems.

Put the complex number z = eπi in the “Cartesian” form z = a+ bi.

Solution. z = eπi = 1(cos(π) + i sin(π)) = 1(−1 + 0i) = −1

It seems remarkable that if we combine the three strangest mathconstants, e, i and π we get

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.

Put the complex number z = 2e13π6i in the “Cartesian” form z = a+ bi.

Solution.z = 2e

13π6i = 2 cos(13π6 ) + 2i sin(13π6 ) = 2 cos(π6 ) + 2i sin(π6 ) =

√3 + i.

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Some worked examples.

Put the complex number z = 18 + 26i in the “polar” form z = reiθ wherer, 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

θ = arctan(2618). (The angle θ is about 0.96525166319.)

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Some worked examples.

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 forthis cube root.

Using the previous problem, we write z = 18 + 26i =√103 eiθ where

θ = arctan(2618).

The cube root of√103 eiθ is

√10 ei

θ3

(The angle θ3 is about 0.3217505544.) Using a calculator, we can see that

this comes out to approximately

√10 cos(

θ

3) + i√10 sin(

θ

3) = 3 + i.

One could check by computing (3 + i)3 and see that we indeed get18 + 26i.

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Some worked examples.

Find a complex number z such that ln(−1) = z.

Solutions. Since −1 in polar coordinate form is −1 = eiπ then z = πi is asolution to ln(−1).

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Some worked examples.

A question found on the internet: What is ii?

We can find one answer if we write the base i in polar form i = eπ2i.

(More carefully, we might note that i = eπ2i+2πki, for any integer k.)

Then ii = (eπ2i)i = e

π2i2 = e−

π2

≈0.207879576350761908546955619834978770033877841631769608075135...

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Complex numbers v. Real numbers

Here are some things one can do with the real numbers:

1 Show that f(x) = sinx 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).

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).

These are all topics for further exploration in a course in complex variables.Smith (SHSU) Elementary Functions 2013 13 / 14

Last Slide!

It is appropriate that we end our series of precalculus lectures with apresentation of Euler’s marvelous formula, which brings together both thetrigonometric 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!

eiθ = cos θ + i sin θ

(End)

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