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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 + 201 = (—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 : zei6, 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 : Z3€3w : 20ei¢. 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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 Fall '05
 ZUCKER
 Calculus, Complex Numbers

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