Complex Numbers PDF

1. Complex Numbers Complex numbers were introduced by Gerolamo Cardano to solve certain cubic equations. He used them as a tool to find a closed formed solutions for such equations without possibly realizing that this tool will prove to be his most important con...

1. Complex Numbers Complex numbers were introduced by Gerolamo Cardano to solve certain cubic equations. He used them as a tool to find a closed formed solutions for such equations without possibly realizing that this tool will prove to be his most important contribution to mathematics. One way to motivate complex numbers is to consider the polynomial x2 + 1. It has no real roots as the square of a real number is always non-negative. So mathematicians said that let’s imagine that there is a ”number” i which has exactly one property that i2 = −1. It turns out that if we add i to the set of real numbers and try to perform usual arithmetic operations, we get a larger set of numbers, which we denote C, that behaves very much like R. Although, it should be emphasized that the similarity is limited to basic arithmetic operations and in many ways, complex numbers are different from real numbers. For instance, we can get absurd conclusions as below if we are not careful: r √ √ r −1 1 −1 = −1 so = 1 −1 √ √ −1 1 so √ = √ 1 −1 √ √ √ √ cross multiplying −1 −1 = 1 1 =⇒ −1 = 1 As some of you may have guessed correctly, the issue is with the step r √ 1 1 =√ −1 −1 √ √ √ If they were equal, then by cross multiplication, we would have −1 −1 = 1, which is absurd. √ This surprises us because we are conditioned to view the radical as a function. This is true for non-negative real numbers, provided we work with the convention that square root of a positive number is defined to be its positive root. With this convention, we indeed have a function √ : R+ ⊔ {0} → R+ ⊔ {0}. But the moment we try to extend this function to the larger domain √ Is there a function like this? : R −−−−−−−−−−−−−−−→ C we land into several issues. The earlier convention ”take the positive root” doesn’t work any √ √ √ more as i is neither positive nor negative. We will simply state that formulas like xy = x y don’t work for complex numbers. Apart from such issues, the arithmetic properties of complex number (i.e. the 4 fundamental arithmetic operations) behave exactly the way they do for real numbers with an additional relation i2 = −1. 1 2 Before we discuss further about complex numbers, a bit of motivation. They are not just a mathematical tool to solve algebraic equations. In fact, complex numbers have found ap- plications in diverse areas like electrical engineering (signal processing, AC circuits, power distribution), quantum mechanics and fluid dynamics! 1.1. Complex number as an ordered pair in Argand plane. We represent a complex number z ∈ C as z = x + iy where x, y ∈ R. Take the xy-plane and replace the x-axes and y-axes with real and imaginary lines, respectively. We say that x is the real part of z and write x = Re z. We say that y is the imaginary part and write y = Im z. Here iy means simply this that when we take the square of it, we get −y 2. That is the algebraic meaning of iy. Later on we will see a geometric interpretation of multiplication by i. Two important operations on a complex numbers are (1) Conjugation: C → C defined as z = x + iy 7→ z = x − iy. This induces a bijection of a complex numbers and as you will later learn, it preserves underlying algebraic and geometric properties. p (2) Modulus (or absolute value): C → R+ ∪ {0} defined as z = x + iy 7→ |z| = x2 + y 2. Few properties: (1) |z| ≥ 0 and |z| = 0 ⇐⇒ z = 0. (2) |z| = |z|. (3) zz = |z|2. (4) z = z ⇐⇒ z ∈ R. (5) z = −z ⇐⇒ z is purely imaginary i.e. Re(z) = 0. Conjugation commutes with usual arithmetic operations: (1) z ± w = z ± w. (2) zw = z w. z z (3) ( ) =. w w Modulus behaves well with respect to ONLY multiplication and division: (1) |zw| = |z| |w|. z |z| (2) =. w |w| It is not well behaved with respect to addition or subtraction. For example, putting z = 1, w = −1, we see that 0 = |z + w| = ̸ |z| + |w| = 2. Complex number as a vector in R2 plane: Sometimes, we also view z = x + iy as the vector joining origin to the point (x, y) ∈ R2. In that case, |z| is the magnitude of that vector, which is the precisely the length of the line joining origin to the point (x, y). The direction of this vector is defined as (a convention) to be the angle that it makes with x-axes measured anticlockwise. This interpretation makes sense ONLY if z ̸= 0.  3 Note that if |z| = r and the angle that z makes from the x-axes is θ then x = r cos θ and y = r sin θ. So we can also write any non-zero complex number uniquely as z = r(cos θ+i sin θ). This is called the polar form of a complex number. The angle θ is the argument of z. Exercise 1.1. Express the following complex numbers in polar form with argument in [0, 2π). (1) 2 + 3i. (2) 2 − 3i. (3) −2 + 3i. (4) −2 − 3i. √ Hint: Each of this complex number lies on a circle with radius 13 (22 + 32 = 13). Find 3 θ ∈ [0, π/2] such that sin(θ) = √. Now work out arguments of the remaining 3 numbers. 13 The exercise (1.1) reveals the geometric meaning of multiplication by i. If z = x + iy then iz = −y + ix. Note that |z| = |iz|. Thus, if we represent z on the argand plane, then |z| and |iz| are both on a circle centered at origin with radius |z|. One can quickly check using coordinate geometry ”slope of the line method”, that iz is a π/2 rotation of z in anticlockwise direction from the x-axes. A direct (and more elementary) proof is clear from the following picture: Figure 1. Geometric meaning of multiplication by i. Remark Few points: (1) Any identity or relation involving complex numbers is preserved if we apply conjugation i.e. under the transformation i 7→ −i. Thus, if for two complex numbers z and w, zw + w = 1 then applying conjugation gives zw + w = 1. (2) One way to view conjugation is to think of it as a reflection under x-axes as it keeps x component constant but changes the sign of the y-component. Let’s apply this theory to solve an example 4 Example 1.1. Suppose two complex numbers satisfy |z|2 w − |w|2 z = z − w. Can we break this relation in simpler form? We use zz = |z|2 and so on to write zzw − wwz = z − w =⇒ zw(z − w) = z − w. Rearranging the terms, we get z(1 − zw) = w(1 − wz). Put zw = α. Clearly α = 1 =⇒ the above equality. So suppose α ̸= 1. Taking conjugates, we get z(1 − wz) = w(1 − zw). Multiplying the two equations, and dividing on both sides by (1 − α)(1 − α), we get |z| = |w|. Putting this in the original equation, we get (|z|2 + 1)(z − w) = 0. Thus, z = w. So |z|2 w − |w|2 z = z − w ⇐⇒ either zw = 1 or w = z. Review questions: Exercise 1.2. It is a standard fact that if z is a complex root of a quadratic polynomial ax2 + bx + c where a, b, c ∈ R, then z is also a root. This observation is true more generally: Let f (x) be a polynomial with coefficients in R. Suppose z ∈ C be a root of f. Show that z is also a root of f. Do we really need coefficients to be in R ? Exercise 1.3. Using properties of complex numbers, show that a cubic polynomial with real coefficients always has a real root. Exercise 1.4. Find the value of in + in+1 + in+2 + in+3 for any n. Exercise 1.5. Let z be a complex number lying on the unit circle i.e. |z| = 1. Show that arg z |1 − z| = 2 sin. 2 1.2. Euler’s formula. Euler, based on Madhava’s series expansion of trigonometric functions, observed the following formula: eix = cos x + i sin x, for any x ∈ R. (1) A special case of the Euler’s formula is eiπ +1 = 0. It connects fundamental constants in math- ematics 1, π, 0, i and e, written in the chronological order of their introduction to mathematics as symbols!  5 1.3. A geometric argument to suggest Euler’s formula. : One can always derive Euler’s formula using the series expansion of trigonometric functions. But a more geometric approach is stated below. Let y = f (x) be a differentiable curve C in the R2 plane. Assume that for any point (x0 , y0 ) ∈ C, the tangent line is perpendicular to the line joining origin to the point (x0 , y0 ). y0 This means that f ′ (y0 ) · = −1. Since this holds for every point (x, y) ∈ C, we get the x0 dy x following differential equation = −. Solving this equation gives that x2 + y 2 = λ for some dx y constant λ. This shows that any such curve y = f (x) must be a circle centered at origin! If further, f (0) = 1, then we get that λ = 1 and it must be the unit circle. Now let’s look at the equation v = f (t) = eit. Throwing caution to the wind, and assuming that it can be done, we differentiate this equation with respect to t to get f ′ (t) = iv. We have already seen that multiplication by i means rotating the vector v by an angle of π/2 around origin in anticlockwise direction. Thus, w = f (t) traces a curve in argand plane such that tangent at any point is perpendicular to the line joining that point to origin (precisely the vector w!). This means that f (t) must lie on a circle centered at origin. Further, f (0) = e0 = 1. ∂f (t) Thus, it must be a unit circle. Further anylysis, by computing the velocity. suggests an ∂t expression for eit as cos t + i sin t. 1.4. Some applications of Euler’s formula. Theorem 1.1 (DeMoivre’s Theorem). For any θ ∈ R and n ∈ N, (cos θ + i sin θ)n = cos nθ + i sin nθ. Putting n = 2, deduce that sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ. Exercise 1.6. If z = r(cos θ + i sin θ), find z̄ (in polar coordinates). If w = s(cos ψ + i sin ψ), then find zw in polar coordinates. If you do the previous exercise, you will notice that the product of complex numbers z and w is a complex number u such that |u| = |z||w| and arg u = arg z + arg w. Similarly, the quotient z/w is a complex number with absolute value |z|/|w| and argument arg z − arg w. 1.5. Complex numbers as matrices. We saw earlier that i : C → C is a linear transforma- tion, which takes a complex number z as a vector and rotates it around the origin by π/2. We take this as a motivation to discuss rotation about origin. Question: We want to find a map Tθ : C → C which rotates every non-zero z by an angle θ around the origin in anticlockwise direction. This question is better to phrase in polar coordinates. Suppose z = reiα. Then the coordinates of Tθ (z) are rei(α+θ). Thus, our map Tθ acts as below: (r cos α, r sin α) 7→ (r(cos α cos θ − sin θ sin α), r(sin α cos θ + cos α sin θ)). 6 This cumbersome formula can be restated more simply in matrix form as    cos θ − sin θ x (x, y) 7→. sin θ cos θ y   cos θ − sin θ We call the matrix the rotation matrix and it can be identified with the sin θ cos θ complex number eiθ as they both have same effect. Thus, we see that with usual addition and multiplication, we can identify the complex num- bers with a subset of 2 × 2 matrices, where under the identification   a −b z = a + bi 7→. b a Note that 0 goes to the zero matrix. Exercise 1.7. Check that complex addition and multiplication are satisfied.   0 −1 Exercise 1.8. We note that i =. Verify that the square of this matrix is −I2×2. 1 0 The complex number operations - modulus and conjugation have matrix analogues. Can you guess them? Determinants and transpose! So if you feel uncomfortable about a number whose square is −1, you can work with a 2 × 2 matrix whose square is −I2×2.