Complex Numbers, Trig Functions and Roots of 1Date: 10/30/97 at 22:00:19 From: Jonah Knobler Subject: TRIG FUNCTIONS AND ROOTS OF 1 Hi! I am a high school junior in precalculus. We just began studying the trigonometric functions two days ago in class. I took special note of the fact that the sin(pi/4) = cos(pi/4) = sqrt(2)/2. This struck a familiar chord because I know (just as a trivial fact) that the square root of i (and therefore the fourth root of -1) is [sqrt(2)/2] + i[sqrt(2)/2]. Representing a point on the unit circle as (cos x, sin x), you can represent the point at the end of an arc with length pi/4 radians as [sqrt(2)/2, sqrt(2)/2] on the cartesian plane. But if you instead use the complex plane (with real and imaginary axes), you get [sqrt(2)/2, i*sqrt(2)/2] - which is exactly the square root of i! This seemed exciting and thought provoking. I realized that this complex number is also the EIGHTH ROOT of 1. Then, it struck me that this point is exactly 1/8 of the way around the UNIT circle, with radius 1! I tried the sin and cos of an arc length whose terminal point was 1/6 of the way around the unit circle, represented it as a complex number, and this was the SIXTH root of 1! The more I tried, the more I became convinced that, for a an arc of length x in radians: (cos x + i sin x)^(2 pi / x) = 1. The base of the above is the point in trig notation, and the exponent is the reciprocal of its distance around the unit circle. This has worked for every sine and cosine pair I have tried. It's cool but why does it work? I have thought about it for a long time and can't think of what in the world trig functions (i.e. ratios of triangle sides) have to do with imaginary numbers! Please, is this true, and why? Also, if you didn't use a UNIT circle, what would happen? In other words, if you represented the terminal point of an arc of length pi/4 radians on a circle of radius 2 (instead of 1) as a complex number, would these coordinates be the 8th root of 2 (instead of 1)? Or some other related number? This is cool but confusing! Please explain... -Jonah Knobler Date: 10/31/97 at 08:13:19 From: Doctor Anthony Subject: Re: TRIG FUNCTIONS AND ROOTS OF 1 You have chanced upon some properties of complex numbers with which you will become familiar when you study the subject in detail. A complex number is in general represented by z = x + i.y where x is the real component and y the imaginary component. The x axis is therefore called the real axis and the y axis the imaginary axis. If you express these components in polar coordinates with x = r.cos(theta) and y = r.sin(theta) we have: z = r(cos(theta) + i.sin(theta)) Now if you square a complex number, you square the value of r and you DOUBLE the angle theta. If you cube a complex number you cube the value of r and TREBLE the angle theta. [This is easily proved when you have done the Addition Theorems in trigonometry, such as sin(A+B) = sin(A)cos(B) + cos(A)sin(B) ] Going in the reverse direction, if you find the square root of a complex number you take the square root of r and HALVE the angle theta. If you have a point on the unit circle with r=1, then squaring or square rooting keeps you on the circle, but you move around it as theta is doubled or halved. Consider the point i on the unit circle. Here z = cos(pi/2) + i.sin(pi/2) Then z^(1/2) = cos(pi/4) + i.sin(pi/4) = 1/sqrt(2) + 1.1/sqrt(2) = sqrt(2)/2 + i.[sqrt(2)/2] If you wanted the 8th roots of 1, you will find that these lie uniformly spaced like spokes of a wheel round the unit circle at theta = 0, pi/4, pi/2, 3.pi/2, pi, 5.pi/4, 3.pi/2, 7.pi/4 The general method analytically to get these points is to write: z = cos(0+2k.pi) + i.sin(0+2k.pi) where k = 0, 1, 2, ....., 7 The 2k.pi simply adds complete revolutions to the initial value of theta. However when you divide the angle by 8, and let k run through values 0 to 7 you obtain 8 different positions round the circle. If you let k take further values 8, 9, 10, ... you get back to positions already found, so there are just eight 8th roots of 1. z^(1/8) = cos(k.pi/4) + i.sin(k.pi/4) With k=0 we get z = cos(0) + i.sin(0) = 1 k=1 z = cos(pi/4) + i.sin(pi/4 = sqrt(2)/2 + i.sqrt(2)/2 k=2 z = cos(pi/2) + i.sin(pi/2) = i ............................................. ............................................. and so on. If you are looking at points which are not on the unit circle, then taking roots will also affect the value of r, and you move to a circle with radius sqrt(r) or r^(1/3) or r^(1/8) or whatever the type of root you are taking. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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