Math 632: Complex Analysis Chapter 1: Complex numbers

Transcription

1 Math 632: Complex Analysis Chapter 1: Complex numbers Spring 2019

2 Definition We define the set of complex numbers C to be the set of all ordered pairs (a, b), where a, b R, and such that addition and multiplication are defined by (a, b)+(c, d) = (a+c, b+d), (a, b) (c, d) = (ac bd, bc+ad). Then C is a field. In other words, C satisfies the usual associative, commutative and distributive laws for addition and multiplication, the identity for addition is (0, 0), the identity for multiplication is (1, 0) and there are additive and multiplicative inverses for each non-zero element in C.

3 Properties of C The mapping R C given by a (a, 0) is a field isomorphism of R into C. This means we can consider R as a subset of C. If we define i = (0, 1), then we can write (a, b) = a + bi. We will use this notation for complex numbers from now on. Now i 2 = 1, which means the equation z = 0 has a root in C. In fact, we can factorize this as z = (z i)(z + i). More generally, if z and w are complex numbers, then z 2 + w 2 = (z + iw)(z iw). Multiplicative inverses of non-zero complex numbers can be calculated to be 1 = a ib ( ) a + bi a 2 + b 2 = a b a 2 + b 2 i a 2 + b 2.

4 Real and imaginary parts Definition Writing z = x + iy C, we call x the real part of z and y the imaginary part of z. We write these as x = Re(z) and y = Im(z). Example Let z = a + ib and w = c + id. Then Re(z + w) = Re(a + c + i(b + d)) = a + c = Re(z) + Re(w). Further, Re(zw) = Re(ac bd + i(bc + ad)) = ac bd = Re(z) Re(w) Im(z) Im(w).

5 Absolute values and conjugates Definition If z = x + iy, then the absolute value of z is z = (x 2 + y 2 ) 1/2 and the conjugate of z is z = x iy. Note that z 2 = zz. Example Let z = 3 + 4i. Then z = 3 4i and z = = 5. Example If z 0, then 1 z = x iy x 2 + y 2 = z z 2.

6 Properties of absolute values and conjugates The following properties are fairly straightforward to prove: Re(z) = (z + z)/2 and Im(z) = (z z)/2i. (z + w) = z + w and zw = z w. zw = z w. z/w = z / w if w 0. z = z.

7 Algebra and geometry So far we have considered C as an algebraic object, but it is often very useful to consider the geometry of the complex numbers. Every point z = x + iy in C can be identified with a unique point (x, y) R 2. Example If z, w C, draw the straight lines from 0 to z and 0 to w. Then the fourth point needed to complete the parallelogram is z + w. Example The quantity z w is exactly the distance between the two points z and w in the plane with the usual Euclidean distance.

8 The triangle inequality Theorem If z, w C, then z + w z + w. Proof. To prove this, first observe that z Re(z) z, z Im(z) z. Hence Re(zw) zw = z w. Therefore z + w 2 = z Re(zw) + w 2 z z w + w 2 = ( z + w ) 2.

9 Consequences of the triangle inequality When do we have equality in the triangle inequality? We need Re(zw) = zw. This is equivalent to zw 0. So either w = 0, or z/w 0 (why?), which means that z = tw for some t 0. By induction, we obtain z 1 + z z n z 1 + z z n. From the triangle inequality, we can obtain the reverse triangle inequality, z w z w. Geometrically speaking, absolute value is the notion of distance. Also geometrically speaking, taking a complex conjugate corresponds to a reflection in the real axis (i.e. if z = x + iy C, the line y = 0).

10 Polar coordinates Given z = x + iy C, we can consider the polar coordinates (r, θ) where x = r cos θ and y = r sin θ. We must have that r 0 is the absolute value z, θ is the angle that the line between 0 and z makes with the positive real axis. Note that θ is not defined for z = 0, and that θ is only defined up to integer multiples of 2π. The angle θ is called the argument of z, denoted arg(z), and is not a function because of the ambiguity of θ. Definition We define e iθ = cos θ + i sin θ. Note that the book uses the notation cis θ. Hence z = z e i arg(z).

11 Examples Example If z > 0, then θ = 0 and r = z. If z < 0, then θ = π and z = z e iπ. In particular, e iπ + 1 = 0. If z = 2i, then θ = π/2 and r = 2 so z = 2e iπ/2. If z = 1 i, then θ = π/4 and r = 2, so z = 2e iπ/4. If z = 3e 7iπ/3, then z = 3 cos(7π/3) + 3i sin(7π/3) = 3 cos(π/3) + 3i sin(π/3) = i 2.

12 The Principal Argument Definition Given z C \ {0}, its principal argument is the argument of z contained in the interval ( π, π], and written as Arg(z). Clearly Arg(z) is an argument of z, but we have removed the ambiguity in its definition by specifying the range to be in a given interval. There is more than one way to specify a principal argument, e.g. the range could be [0, 2π), but we will stick with ( π, π]. An advantage is that now Arg : C \ {0} ( π, π] is a function. A disadvantage is that the function is not continuous because of the negative real axis.

13 Properties of e iθ Let z = re iθ and w = se iφ. Then zw = rs[(cos θ cos φ sin θ sin φ)+i(sin θ cos φ+sin φ cos θ)]. Using the addition formulas for trig functions, we obtain zw = rse i(θ+φ). In particular, arg(zw) = arg(z) + arg(w). By induction, z 1 z 2... z n = r 1 r 2... r n e i(θ θ n). In particular, z n = r n e inθ holds for all integers. When r = 1, we get de Moivre s formula (cos θ + i sin θ) n = cos nθ + i sin nθ.

14 Roots of unity Given a complex number w 0 and a positive integer n, how many solutions to z n = w are there? Suppose that w = se iφ. Then it is not too hard to find a solution: z = s 1/n e iφ/n works. This is not the only solution because z = s 1/n e i(φ+2π)/n also works. In fact, each of the numbers s 1/n e i(φ+2πk)/n, 0 k n 1 is a solution. We call such numbers n th roots of w. If w = 1, these carry the name n th roots of unity. These are all the possible solutions, so there are n distinct n th roots of w.

15 Example Example Find all 4 th roots of w = 4i. First write w in polar form, w = 4e iπ/2. Then the first solution we find is z 0 = 4 1/4 e iπ/8 = 2e iπ/8. The further solutions are z 1 = 2e i(π/2+2π)/4 = 2e 5iπ/8, z 2 = 2e i(π/2+4π)/4 = 2e 9iπ/8, z 3 = 2e i(π/2+6π)/4 = 2e 13iπ/8,

16 Dealing with infinity We will sometimes need to deal with functions that become infinite near a particular point. To deal with such situations, we introduce the extended complex plane C = C { }. How do we treat the point? Stereographic projection: Let S be the unit sphere in R 3, S = {(x 1, x 2, x 3 ) R 3 : x x x 2 3 = 1}, and let N = (0, 0, 1) be the north pole. Identify C with the plane {(x 1, x 2, 0) : x 1, x 2 R}. For each point z in C, draw a straight line in R 3 connecting z and N. This intersects S in exactly one point Z N. If z > 1, then Z is in the northern hemisphere, if z < 1, then Z is in the southern hemisphere and if z = 1 then z = Z. As z, Z approaches N in S. Hence S is a model for C.

17 The Riemann sphere Although we will not say any more about this, S is the first example of a Riemann surface, and is called the Riemann sphere. One can calculate that if z C, then Z = ( z + z i(z z), 1 + z z 2, z z 2 ). Given Z N, we find that z = x 1+ix 2 1 x 3. A distance function d can be put on C using the Euclidean distance on S. If z, w C, this is d (z, w) = 2 z w (1 + z 2 )(1 + w 2 ), and if z C, d (z, ) = z 2.