Eulers Equation The value of complex numbers was recognized but - PowerPoint PPT Presentation

Euler’s Equation The value of complex numbers was recognized but poorly understood Elementary Functions 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. The complex plane (“Argand diagram” or “Gauss plane”) was introduced Dr. Ken W. Smith 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 Sam Houston State University around the same time. 2013 Smith (SHSU) Elementary Functions 2013 1 / 14 Smith (SHSU) Elementary Functions 2013 2 / 14 Euler’s Equation Euler’s Equation Notice the following remarkable fact that if √ 2 + 1 3 2 i = cos π 6 + i sin π Euler would explain why that was true. Using the derivative and infinite z = series, he would show that 6 then z 3 = i. (Multiply it out & see!) Thus z 12 = 1 and so z is a twelfth e iθ = cos θ + i sin θ (2) root of 1. Now the polar coordinate form for z is r = 1 , θ = π 6 , that is, z is exactly By simple laws of exponents, ( e iz ) n = e inz and so Euler’s equation 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! 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 6 + i sin π 6 which is one-twelfth of the way around the unit circle; (cos θ + i sin θ ) n = 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 2 π : (1 , 0) = 1 + 0 i. 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. Smith (SHSU) Elementary Functions 2013 3 / 14 Smith (SHSU) Elementary Functions 2013 4 / 14

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 e iθ = cos θ + i sin θ e iθ = 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 and then wrote the trig functions in terms of the exponential function! formula for cos z and sin z as a sum of exponential functions. By Euler’s formula, with input − z , cos z = e iz + e − iz e − iz = cos( − z ) + i sin( − z ) = cos( z ) − i sin( z ) . 2 Add the expressions for e iz and e − iz to get e iz + e − iz = 2 cos( z ) sin z = e iz − e − iz 2 i and so cos z = e iz + e − iz . (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 2 i , we have a formula for sine: And conversely, the exponential functions are trig functions! sin z = e iz − e − iz . (4) Smith (SHSU) Elementary Functions 2013 5 / 14 Smith (SHSU) Elementary Functions 2013 6 / 14 2 i 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 = e πi in the “Cartesian” form z = a + bi . 6 i in the “Cartesian” form z = a + bi . Solution. z = e πi = 1(cos( π ) + i sin( π )) = 1( − 1 + 0 i ) = − 1 13 π Put the complex number z = 2 e It seems remarkable that if we combine the three strangest math Solution. √ 6 i = 2 cos( 13 π 13 π 6 ) + 2 i sin( 13 π 6 ) = 2 cos( π 6 ) + 2 i sin( π constants, e, i and π we get z = 2 e 6 ) = 3 + i. e πi = − 1 . Some rewrite this in the form e πi + 1 = 0 (often seen on t-shirts for engineering clubs or math clubs.) Smith (SHSU) Elementary Functions 2013 7 / 14 Smith (SHSU) Elementary Functions 2013 8 / 14

Some worked examples. Some worked examples. Find a cube root of the number z = 18 + 26 i and put this cube root in the “Cartesian” form z = a + bi . (Use a calculator and get an exact value for this cube root. √ 10 3 e iθ where Using the previous problem, we write z = 18 + 26 i = Put the complex number z = 18 + 26 i in the “polar” form z = re iθ where θ = arctan( 26 18 ) . r, θ ∈ R and both r and θ are positive. √ 10 3 e iθ is The cube root of Solution. The modulus of z = 18 + 26 i is 18 2 + 26 2 = 1000 . √ 10 e i θ √ 3 10 3 e iθ where So the polar coordinate form of z = 18 + 26 i is θ = arctan( 26 18 ) . (The angle θ is about 0 . 96525166319 .) (The angle θ 3 is about 0 . 3217505544 .) Using a calculator, we can see that this comes out to approximately √ 10 cos( θ √ 10 sin( θ 3) + i 3) = 3 + i. One could check by computing (3 + i ) 3 and see that we indeed get 18 + 26 i . 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 i i ? π 2 i . We can find one answer if we write the base i in polar form i = e Find a complex number z such that ln( − 1) = z. π 2 i +2 πki , for any integer k .) Solutions. Since − 1 in polar coordinate form is − 1 = e iπ then z = πi is a (More carefully, we might note that i = e solution to ln( − 1) . 2 i 2 = Then i i = ( e 2 i ) i = e π π e − π 2 ≈ 0 . 207879576350761908546955619834978770033877841631769608075135 ... Smith (SHSU) Elementary Functions 2013 11 / 14 Smith (SHSU) Elementary Functions 2013 12 / 14

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 2 π , that is, f ( x + 2 π ) = f ( x ) . It is appropriate that we end our series of precalculus lectures with a presentation of Euler’s marvelous formula, which brings together both the 2 Find an infinite set of numbers, x , such that sin( x ) = 1 / 2 . trigonometric functions and the exponential functions into one form! 3 Find a number x such that e x = 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: e iθ = cos θ + i sin θ 1 Show that f ( x ) = e x is periodic with period 2 πi , that is, f ( x + 2 πi ) = f ( x ) . 2 Find an infinite set of numbers, x , such that e x = 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. Smith (SHSU) Elementary Functions 2013 13 / 14 Smith (SHSU) Elementary Functions 2013 14 / 14