Complex Numbers PDF

Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : [email protected] 1...

Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : [email protected] 1 Module-2: Stereographic Projection 1 Euler’s Formula By assuming that the infinite series expansion x2 x3 ex = 1 + x + + +.... 2! 3! of elementary calculus holds when x = iθ, we can arrive at eiθ = cos θ + i sin θ, which is called Euler’s formula. In general, we define ez = ex+iy = ex.eiy = ex (cos y + i sin y). The n-th Root of Unity The solutions of the equation z n = 1 where n is a positive integer are called the n-th root of unity and are given by 2kπ 2kπ 2kπi z = cos + i sin =e n , n n 2πi k = 0, 1, 2,..., n − 1. If we put ω = cos 2π n + i sin 2π n =e n , the n roots are 1, ω, ω 2 ,... , ω n−1. Geometrically they represent the n vertices of a regular polygon of n sides inscribed in a circle of radius one having center at the origin. The circle has the equation | z |= 1 and is often called the unit circle. 2 Point at Infinity The linear transformation z → w = f (z), where f (z) = λz + µ, λ 6= 0, 2 is a one-one mapping of the finite complex plane onto itself. This is not true of the inversion map z → w = 1/z. Writing into polar forms, we have z = reiθ and w = ρeiφ , where ρ = 1/r. Therefore, the points close to the origin in the z-plane, r ≈ 0, are mapped onto points far away from the origin in the w-plane. All the points inside a disk of small radius ε, in the z-plane, are mapped onto points outside a disk of large radius 1/ε, in the w-plane. As ε → 0, the disk in the z-plane shrinks to the origin and there is no image of z = 0 in the w-plane. Similarly, as the point z moves farther and farther away from the origin, its image in the w-plane moves closer and closer to the origin in the w-plane, but there is no point in the z-plane which can be assigned w = 0 as the image under inversion. It turns out to be useful to introduce the concept of a point at infinity, or z = ∞, as a formal image of z = 0 under the inversion map w = 1/z. The point z = 0 can then be regarded as the image of the point at infinity. The use of z = ∞ will always be understood in terms of a limiting process w → 0, where w = 1/z. To examine the 1 behavior of f (z) at z = ∞, it suffices to let z = w and examine the behavior of f ( w1 ) at 2z−1 w = 0. For example, we say that the function f (z) = z−3 tends to 2 as z → ∞, because 2−w f (1/w) = 1−3w tends to 2 as w → 0. 3 Extended Complex Plane By the extended complex number system, we shall mean the complex plane C together with a symbol ∞ which satisfies the following properties : (a) If z ∈ C, then we have z + ∞ = z − ∞ = ∞ , z/∞ = 0. (b) If z ∈ C, but z 6= 0, then z.∞ = ∞ and z/0 = ∞. (c) ∞ + ∞ = ∞.∞ = ∞ (d) ∞/z = ∞ (z 6= ∞). The set C ∪ {∞} is called the extended complex plane and is denoted by C∞. The nature of Argand plane at the point at infinity is made much clear by the use of Riemann’s spherical representation of complex numbers, which depends on Stereographic Projection. 3 Stereographic Projection We consider the Argand plane C and a unit sphere Ŝ tangent to C at z = 0. The diameter NS is perpendicular to C and we call the points N and S the north and south poles of Ŝ respectively. Now we establish a one-one correspondence between the points on the sphere and the points on the plane. To each point A in the plane there corresponds a unique point A0 on the sphere. The point A0 is the point where the line joining A to the north pole N intersects the sphere. Conversely, corresponding to each point A0 on the sphere (except the north pole) there exists a unique point A in the plane. By defining that the north pole N corresponds to the point at infinity, we can say that there exists a one-one correspondence between the points on the sphere and the points in the extended complex plane. This sphere is known as Riemann sphere and the correspondence is known as stereographic projection, (see Fig. 1.1). We consider the sphere as Fig. 1.1: x21 + x22 + x23 = 1, the plane of projection as x3 = 0 and let (0, 0, 1) be the coordinate of N. For any point 4 A0 = (x1 , x2 , x3 ) on the sphere we have point A = (x, y, 0) in the x3 -plane where the line NA0 meets the plane of projection. Obviously, the points (0, 0, 1), (x1 , x2 , x3 ) and (x, y, 0) are collinear and the equation of the line is x1 x2 x3 − 1 = =. x y −1 From this we get x1 x2 x = , y =. 1 − x3 1 − x3 x1 +ix2 Therefore z = x + iy = 1−x3. From this we get x21 + x22 1 + x3 | z |2 = 2 = , (1 − x3 ) 1 − x3 and hence | z |2 −1 x3 =. | z |2 +1 Also we see that z+z z−z 2 = x1 and = x2. | z | +1 i(| z |2 +1) In this way we can establish an one-one correspondence between the points in the extended complex plane and points on the sphere. Example 1.1. Find all the roots of the equation z 4 − (1 − z)4 = 0. z Solution. Let w = 1−z. Then the given equation becomes w4 = 1 = cos 2kπ + i sin 2kπ, where k is an integer. Therefore w = cos 2kπ 4 +i sin 2kπ 4 , k = 0, 1, 2, 3. Again from w = z 1−z 2kπ 2kπi w cos +i sin 2kπ e 4 we get z = w+1. Hence z = cos 2kπ 4 +i sin 2kπ 4 +1 = 2kπi , k = 0, 1, 2, 3. 4 4 e 4 +1 Example 1.2. Find all the values of z for which z 5 = 32 and locate these values in the Argand plane. Solution. z 5 = 32 = 32(cos 2kπ + i sin 2kπ), k = 0, ±1, ±2,... This gives      2kπ 2kπ z = 2 cos + i sin , k = 0, 1, 2, 3, 4. 5 5 5 If k = 0, then z = z1 = 2[cos 0 + i sin 0] = 2. If k = 1, then z = z2 = 2[cos 2π 5 + i sin 2π 5 ]. If k = 2, then z = z3 = 2[cos 4π 5 + i sin 4π 5 ]. If k = 3, then z = z4 = 2[cos 6π 5 + i sin 6π 5 ]. If k = 4, then z = z5 = 2[cos 8π 5 + i sin 8π 5 ]. These are the only roots of the given equation. The values of z are indicated in the figure (see Fig. 1.2). Note that they are equally spaced along the circumference of a circle having center at the origin and the radius 2. Another way of saying this is that the roots are represented by the vertices of a regular polygon. Fig. 1.2: √ 1 Example 1.3. Find all the roots of (−8 − 8 3i) 4 and exhibit them geometrically. Solution.  41 √ 1      4π 4π (−8 − 8 3i) 4 = 16 cos 2kπ + + i sin 2kπ + 3 3   4π   4π  2kπ + 3 2kπ + 3 = 2 cos + i sin , 4 4 k = 0, 1, 2, 3. Therefore all the four roots are z1 = 2(cos π/3 + i sin π/3); z2 = 2(cos 5π/6 + i sin 5π/6); z3 = 2(cos 4π/3 + i sin 4π/3); z4 = 2(cos π/6 − i sin π/6); √ √ √ √ or 1 + i 3, − 3 + i, −1 − i 3, 3 − i. 6  The roots lie at the vertices of a square inscribed in a circle of radius 2 centered at the origin, also are equally spaced with difference of angle π/2 (see Fig. 1.3). Fig. 1.3: Example 1.4. Establish the relation : n−1   n Y kπ = sin , n ≥ 2. 2n−1 k=1 n 2kπi Solution. Let 1, ρ1 , ρ2 ,... , ρn−1 be the n roots of unity, where ρk = e n , k = 1, 2,..., n − 1. Then z n − 1 = (z − 1)(z − ρ1 )(z − ρ2 )...(z − ρn ). Dividing both sides by z − 1 and letting z → 1, we obtain n = (1 − ρ1 )(1 − ρ2 )...(1 − ρn−1 ). Taking conjugate of both sides, we obtain n = (1 − ρ1 )(1 − ρ2 )...(1 − ρn−1 ). 7 Therefore n−1 Y 2 n = (1 − ρk )(1 − ρk ) k=1 n−1 2kπi 2kπi Y = (1 − e n )(1 − e− n ) k=1 n−1   Y 2kπ = 2 1 − cos k=1 n n−1   Y 2 kπ = 4 sin k=1 n n−1   2(n−1) Y 2 kπ = 2 sin. k=1 n Taking the nonnegative square root of both sides we obtain the required result. Example 1.5. For any two nonzero complex numbers z1 and z2 prove that z1 z2 | z1 + z2 | | + | ≤ 2(| z1 | + | z2 |). | z1 | | z2 | Solution. We have z1 z2 | z1 + z2 | | + | | z1 | | z2 | z1 | z2 | +z2 | z1 | = | z1 + z2 | | | | z1 || z2 | | (z1 | z2 | +z2 | z1 |) | = | z1 + z2 | | z1 z2 | | z1 + z2 | ≤ (| z1 | z2 || + | z2 | z1 ||) | z1 z2 | | z1 + z2 | = 2 | z1 z2 | | z1 z2 | = 2 | z1 + z2 | ≤ 2(| z1 | + | z2 |). 8