5.7 Euler's Marvelous Formula (Slides

Elementary Functions The value of complex numbers was recognized but poorly understood during the late Renaissance period (1500-1700 AD.) The number system Part 5, Advanced Trigonometry was explicitly studied in the late 18th century. Euler used i for the square Lecture 5.7a, Euler’s Marvelous Formula root of −1 in 1779. Gauss used the term “complex” in the early 1800’s.

Dr. Ken W. Smith 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 Sam Houston State University doctoral dissertation of Gauss in 1799 and in work of Caspar Wessel around the same time. 2013

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Notice the following remarkable fact that if √ 3 1 π π Euler would explain why that was true. Using the derivative and inﬁnite z = + i = cos + i sin 2 2 6 6 series, he would show that then z3 = i. (Multiply it out & see!) Thus z12 = 1 and so z is a twelfth eiθ = cos θ + i sin θ (2) root of 1. π Now the polar coordinate form for z is r = 1, θ = 6 , that is, z is exactly one-twelfth of the way around the unit circle. z is a twelfth root of 1 and it By simple laws of exponents, (eiz)n = einz and so Euler’s equation is one-twelfth of the way around the unit circle. This is not a coincidence! explains DeMoivre formula. DeMoivre apparently noticed this and proved (by induction, using sum of angles formulas) that if n is an integer then This explains the “coincidence” we noticed with the complex number z = cos π + i sin π which is one-twelfth of the way around the unit circle; n 6 6 (cos θ + i sin θ) = cos nθ + i sin nθ. (1) 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. 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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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 eiθ = cos θ + i sin θ 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. and then wrote the trig functions in terms of the exponential function! By Euler’s formula, with input −z, eiz + e−iz e−iz = cos(−z) + i sin(−z) = cos(z) − i sin(z). cos z = 2 Add the expressions for eiz and e−iz to get eiz + e−iz = 2 cos(z) eiz − e−iz sin z = and so 2i eiz + e−iz cos z = . (3) 2 The exponential and trig functions are very closely related. Trig functions If we subtract the equation e−iz = cos z − i sin z from Euler’s equation are, in some sense, really exponential functions in disguise! and then divide by 2i, we have a formula for sine: And conversely, the exponential functions are trig functions! eiz − e−iz Smith (SHSU) sinElementaryz = Functions . 2013(4) 5 / 14 Smith (SHSU) Elementary Functions 2013 6 / 14 2i Some worked examples. 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.

πi 13π i Solution. z = e = 1(cos(π) + i sin(π)) = 1(−1 + 0i) = −1 Put the complex number z = 2e 6 in the “Cartesian” form z = a + bi.

It seems remarkable that if we combine the three strangest math Solution. 13π √ 6 i 13π 13π π π constants, e, i and π we get z = 2e = 2 cos( 6 ) + 2i sin( 6 ) = 2 cos( 6 ) + 2i sin( 6 ) = 3 + i. 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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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. √ 3 iθ iθ Using the previous problem, we write z = 18 + 26i = 10 e where Put the complex number z = 18 + 26i in the “polar” form z = re where θ = arctan( 26 ). r, θ ∈ and both r and θ are positive. 18 √ R The cube root of 103 eiθ is 2 2 √ Solution. The modulus of z = 18 + 26i is 18 + 26 = 1000. i θ √ 10 e 3 So the polar coordinate form of z = 18 + 26i is 103 eiθ where θ = arctan( 26 ). (The angle θ is about 0.96525166319.) θ 18 (The angle 3 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. Smith (SHSU) Elementary Functions 2013 9 / 14 Smith (SHSU) Elementary Functions 2013 10 / 14 Some worked examples. Some worked examples.

A question found on the internet: What is ii?

π i Find a complex number z such that ln(−1) = z. We can ﬁnd one answer if we write the base i in polar form i = e 2 .

π i+2πki Solutions. Since −1 in polar coordinate form is −1 = eiπ then z = πi is a (More carefully, we might note that i = e 2 , for any integer k.) solution to ln(−1). i π i i π i2 − π Then i = (e 2 ) = e 2 = e 2

≈ 0.207879576350761908546955619834978770033877841631769608075135...

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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). It is appropriate that we end our series of precalculus lectures with a 2 Find an inﬁnite set of numbers, x, such that sin(x) = 1/2. presentation of Euler’s marvelous formula, which brings together both the trigonometric functions and the exponential functions into one form! 3 Find a number x such that ex = 200. 4 Compute ln(2). 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: iθ 1 Show that f(x) = ex is periodic with period 2πi, that is, e = cos θ + i sin θ f(x + 2πi) = f(x). 2 Find an inﬁnite set of numbers, x, such that ex = 1/2. (End) 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.

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