Complex Numbers - On the benefits of using complex numbers...

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Unformatted text preview: On the benefits of using complex numbers As useful as the real number system is, in allowing us to give good quantitative models of the world, it suffers from certain drawbacks observed centuries ago. From a purely mathematical point of view, we can see that there are polynomial equations with real coeffients that don’t have solutions if we insist on operating within the real number system. And it doesn’t help to restrict one’s attention to polynomials with integer coefficients. We take as fundamental example the polynomial P(:13) : 932 + 1. When m is a real number, x2 + 1 2 1, so our P($) has no real zeros. So we make up the zeros of P($)I Call one i; then the other will be —z'. Then 2'2 = —1. Then we tack this thingz' onto the real numbers as follows: The complex numbers are symbols of the form 2 = :6 + yt, with a" and y real. They are in one—to—one correspondence with the pairs of real numbers (say); we understand that y is getting multiplied by 2'. Thus, the complex numbers form a plane. The real numbers consist of those complex numbers that do not involve i, i.e., where y = 0, so are given as usual as a line, namely the :C—axis. The purely imaginary numbers are those with x = 0, namely the real multiples of t. The rules for addition and multiplication are: (x+yi)+(m’+y’i) = ($+m’)+(y+y’)i (w+yi)($’+y’i) = (m’—yy’)+(wy’+w’y)i, just as you would expect if there were no bugs in this system. There are none. I) Example. (3+i)+(5—2t) =8—l, and (3+l)(5—2t) = 15—i—2l2 = 17—t. It is not hard to verify that the algebra goes just as with the real numbers. We make some of this explicit below. The additive identity is the real number 0, and the multiplicative identity is the real number 1. Additive luverses: if z = x + yt, then —2 : (—x) —|— (—y)t, also written —a: — yl. Complex conjugates: if z = m —|— gt, 2 = m — yl (the reflection of 2 through the real—axis in the complex plane picture). Thus, 2 = 2 if and only if z is real; otherwise, 2 and E form a conjugate pair of complex numbers. It is always true that (z + w) : E + E and (2w) : 3E. The modulus of z is defined to be (23%. It is always a non—negative real number, as 2? = 332 +y2; thus =x/a‘2 —|— y2. We have = 0 if and only if z = 0. When z is real, this is nothing other than the familiar absolute value: =\/ 962. Note that we always have [El = 121, and [27.0] = Multiplicative luverses: this is a little trickieri but not difficult if we have ab— sorbed the notion of modulus. 2*1 : é : é : II) Examples. The conjugate of —1 + 2t is —1 — 22', l — 1 —|— 22'] :\/5, and then (—1 + 20-1 = (—1 — 20/5 : —% — a: The one thing you must discard is the notion of inequality between general complex numbers. By this, we mean that there is no notion of inequality that has the same rules of manipulation that inequality of real numbers satisfies. 1 You may be thinking “so what?” but be patient. One thing you may have seen before: let a22+bz+c : 0 (a, b, 0 real, a 7E 0) be the general real quadratic equation. The roots of this are given by the quadratic formula: —b ::V b2 — 46“: 2a 2: When ()2 —4ac < 0, there are no real solutions, but when we allow the larger complex number system, we see a pair of complex conjugate solutions. Conversely, a pair of complex conjugate numbers satisfies a quadratic equation with real coefficients. III) Example. —1 —|— 22' and —1 — 21' are clearly the roots of the equation (2 — (—1 —|— — (—1 — 22') = 0. We can rewrite this as + 1] — + 1] + 22') = O. This is the same as [,2 + 1]2 + 4 = Z2 + 22 + 5 = 0, a quadratic with real coefficients. Next, let P be a polynomial with real coefficients of arbitrary degree. Then P(Z) = Why? It follows that the zeros of P are either real, or come in conjugate pairs, as P(z) = 0 implies P(§) = 0. It follows that P factors (at least, in the abstract) into a product of linear and quadratic factors with real coefficients. (This fact underlies the method of Ch. 8.5, which you are invited to look at.) By the theory of equations, the total number of complex zeros, counted with multiplicity, equals the degree of P. We now connect up with what we did in Week 2. Since we have polar coordinates in the plane, it gives an alternative way of writing complex numbers. If [73 6] is a polar coordinate pair for (3:, y), we know that xzrcosd, yzrsind. In other words, 2 = a: + yz’ can be written as 2 = 7"(c0s 6 +z'si116) (the polar form of the complex number), and = r. (Note that |c0s6 —|— 2' sin 6| 2 1.) The complex numbers of modulus 1 comprise the unit circle; they are the numbers cos 6 —|— 2' sin 6. If it is still “so what?” here comes the main point. You should be able to understand it better after we study power series near the end of the semester. There is the exponential function of a complex variable 63, for which the usual laws of exponents still hold, such that: i) when 2 is real, it is the familiar exponential function, ii) when z 2 29 is purely imaginary ((9 real), (*) em = cos 6 —|— ism 6. You are expected to write down the expression 619 without flinching, and know that it means We have that the polar form of a complex number is given as z : |z|ei6, with 6 determined up to a multiple of 271' (except when 2 = 0). Note that for Z = 620, E = 2—1 = 6—10. IV) Examples- 3) emf/3 : (305(77/3) + isin(7T/3) : % + Also 6” = —1. Cute, eh? And 62+” 2 626” : e2(_1) 2 _e2_ b) Continuing with the last item in a), we have emflflt = 62% You’ll need things like this for Week 4. c) How about writing Z = 1 —i in polar form? Well, =\/§, so 2 =\/§ew for some 6. Which? One for which 6w 2 Thus cos6 = % and sinQ = Basic trigonometry tells you that you may take (9 = —1 (If you draw a picture, this will 4. be clear.) Therefore, ,2 =\/§e_1§. cos 7rt —|— i sin 777i). As we shall see in Week 4, the above enables us to give a unified treatment of the solutions of certain difierential equations. (“Unified” means that we won’t have to divide the discussion into cases.) The polar form supplies a convenient way of taking powers and roots of complex numbers. It’s clear that for any positive integer n, flew)” = rnemg. Let’s see if we can run this backwards, and see the n—th roots of a given complex number. Let 20 = a + bi; we are asking to solve z" = 20. By the theory of equations, there are 71 solutions, and we will see that they all have multiplicity one (except when 20 = 0). To keep things from getting out of hand, let’s take n = 3 for now. The use of rectangular coordinates would have us solving for z = :23 —|— yi by: 23 : — 393342) + i(3:v2y — : a + bi, that is a pair of cubic equations in two unknowns: £83 — 3933f = a, 333% — y3 = b. What do you think of that? { “mick/7’) After writing 20 = lzdem (in polar form), the problem becomes: 23 : |Z|3€3w : |20|ei¢. This equation separates nicely: = |20|%, and 39 = q5 + 2km (16 an integer); 9 = %(q5+2k77) = Up to multiples of 27? again, this gives three possibilities for the terminal side of 6, allowing us to solve the problem efficiently. V) Example. Let’s determine the sixth roots of z = 1 — i. The polar form of this number was calculated in Example IV c): z = 2% €_i%. One of the sixth roots jumps out at us: Zie’ifi. (Recall that (25% = 2%.) The other five are 21—12€i(§—:1T+21€Tfi), with k = 1, 2, 3,4, 5 (k = 6 brings us back to the first one). Perhaps you can see that the n—th roots of a complex number of modulus one consists of n points on the unit circle evenly spaced. Problems. 1. a) Write i3 and i4 in the usual form (a + bi, a, b real). b) Show that for integers n, the value of in depends only on the remainder when n is divided by 4. 2. For 2 = a: + yi, write ,23 in the usual form. 3. Let 2 = 1 — i. Determine E, and 4. Verify the identity: (2w) 2 2E. 5. Solve the equation 22 +\/§z —|— 1 = 0. 6. Write each of the following numbers in polar form (use inverse—trig notation when necessary): —3i, 3 + 4i. 7. Determine the five fifth roots of —3i. 8. Determine all complex numbers other than 2 = —1 that satisfy the equation 23 + 1 = 0 in two ways: a) by using the polar form; b) by the algebra of polynomials (you know, 2 + 1 is a factor of 23 + 1, so How do the results compare? ...
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Complex Numbers - On the benefits of using complex numbers...

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