Complex Numbers and the Complex Exponential

Complex Numbers and the Complex Exponential φ (2+i) i 2 θ φ 2+i θ 1

2 1. Complex numbers The equation x 2 + 1 0 has no solutions, because for any real number x the square x 2 is nonnegative, and so x 2 + 1 can never be less than 1. In spite of this it turns out to be very useful to assume that there is a number i for hich one has (1) i 2 1. Any complex number is then an expression of the form a + bi, here a and b are old-fashioned real numbers. The number a is called the real part of a + bi, and b is called its imaginary part. Traditionally the letters and are used to stand for complex numbers. Since any complex number is specified by to real numbers one can visualie them by plotting a point ith coordinates (a, b) in the plane for a complex number a + bi. The plane in hich one plot these complex numbers is called the Complex plane, or Argand plane. PSfrag replacements b θ arg a Figure 1. A complex number. You can add, multiply and divide complex numbers. Here s ho: To add (subtract) and c + di + (a + bi) + (c + di) (a + c) + (b + d)i, (a + bi) (c + di) (a c) + (b d)i. To multiply and proceed as follos: (a + bi)(c + di) a(c + di) + bi(c + di) ac + adi + bci + bdi 2 (ac bd) + (ad + bc)i here e have use the defining property i 2 1 to get rid of i 2. To divide to complex numbers one alays uses the folloing trick. No a + bi c + di a + bi c + di c di c di (a + bi)(c di) (c + di)(c di) (c + di)(c di) c 2 (di) 2 c 2 d 2 i 2 c 2 + d 2,

3 so a + bi (ac + bd) + (bc ad)i c + di c 2 + d 2 ac + bd bc ad c 2 + + d2 c 2 + d 2 i Obviously you do not ant to memorie this formula: instead you remember the trick, i.e. to divide c + di into a + bi you multiply numerator and denominator ith c di. For any complex number c + di the number c di is called its complex conjugate. Notation: c + di, c di. A frequently used property of the complex conjugate is the folloing formula (2) (c + di)(c di) c 2 (di) 2 c 2 + d 2. 2. Argument and Absolute Value For any given complex number one defines the absolute value or modulus to be a 2 + b 2, so is the distance from the origin to the point in the complex plane (see figure 1). The angle θ is called the argument of the complex number. Notation: arg θ. The argument is defined in an ambiguous ay: it is only defined up to a multiple of 2π. E.g. the argument of 1 could be π, or π, or 3π, or, etc. In general one says arg( 1) π + 2kπ, here k may be any integer. From trigonometry one sees that for any complex number one has so that and a cos θ, and b sin θ, cos θ + i sin θ ( cos θ + i sin θ ). tan θ sin θ cos θ b a. Example: Find argument and absolute value of 2 + i. Solution: 2 2 + 1 5. lies in the first quadrant so its argument θ is an angle beteen 0 and π/2. From tan θ 1 2 e then conclude arg(2 + i) θ arctan 1 2. 3. Geometry of Arithmetic Since e can picture complex numbers as points in the complex plane, e can also try to visualie the arithmetic operations addition and multiplication. To add and one forms the parallelogram ith the origin, and as vertices. The fourth vertex then is +. See figure 2. To understand multiplication e first look at multiplication ith i. If then i i(a + bi) ia + bi 2 ai b b + ai.

PSfrag replacements 4 θ arg c + di + b c + d a PSfrag replacements Figure 2. Addition of and c + di θ arg + c + di c d i(a + bi) b a 90 a + bi b b a Figure 3. Multiplication of a + bi by i. Thus, to form i from the complex number one rotates counterclockise by 90 degrees. See figure 3. If a is any real number, then multiplication of c + di by a gives a ac + adi, so a points in the same direction, but is a times as far aay from the origin. If a < 0 then a points in the opposite direction. See figure 4. Next, to multiply and c + di e rite the product as (a + bi) a + bi. Figure 5 shos a + bi on the right On the left, the complex number is dran, then a as dran. Subsequently i and bi ere constructed, and finally a + bi as dran by adding a and bi. One sees from figure 5 that since i is perpendicular to, the line segment from 0 to bi is perpendicular to the segment from 0 to a. Therefore the larger shaded triangle on the left is a right triangle. The length of the adjacent side is a, and the length of the opposite side is b. The ratio of these to lengths is a : b, hich

PSfrag replacements + c + di c d i(a + bi) b 90 2 bi b θ arg + c + di c d a + bi i(a + bi) b 90 2 Figure 4. Multiplication of a real and a complex number a + bi 2 3 5 2 3 i θ arg ϕ a b a + bi θ arg a Figure 5. Multiplication of to complex numbers is the same as for the shaded right triangle on the right, so e conclude that these to triangles are similar. The triangle on the left is times as large as the triangle on the right. The to angles marked θ are equal. Since is the length of the hypothenuse of the shaded triangle on the left, it is times the hypothenuse of the triangle on the right, i.e.. The argument of is the angle θ + ϕ; since θ arg and ϕ arg e get the folloing to formulas (3) (4) arg() arg + arg, i.e. hen you multiply complex numbers, their lengths get multiplied and their arguments get added. 4. Applications in Trigonometry Unit length complex numbers. For any θ the number cos θ + i sin θ has length 1: it lies on the unit circle. Its argument is arg θ. Conversely, any complex number on the unit circle is of the form cos φ, here φ is its argument.

6 The Addition Formulas. For any to angles θ and φ one can multiply cos θ+ i sin θ and cos φ + i sin φ. The product is a complex number of absolute value 1 1, and ith argument arg() arg + arg θ + φ. So lies on the unit circle and must be cos(θ + φ) + i sin(θ + φ). Thus e have (5) (cos θ + i sin θ)(cos φ + i sin φ) cos(θ + φ) + i sin(θ + φ). By multiplying out the left hand side e get cos(θ + φ) + i sin(θ + φ) cos θ cos φ sin θ sin φ + i(sin θ cos φ + cos θ sin φ) from hich e get the addition formulas for sin and cos: cos(θ + φ) cos θ cos φ sin θ sin φ sin(θ + φ) sin θ cos φ + cos θ sin φ De Moivre s formula. For any complex number the argument of its square 2 is arg( 2 ) arg( ) arg + arg 2 arg. The argument of its cube is arg 3 arg( 2 ) arg() + arg 2 arg + 2 arg 3 arg. Continuing like this one finds that (6) arg n n arg for any integer n. Applying this to cos θ + i sin θ you find that n is a number ith absolute value n n 1 n 1, and argument n arg nθ. Hence n cos nθ+i sin nθ. So e have found (7) (cos θ + i sin θ) n cos nθ + i sin nθ. This is de Moivre s formula. For instance, for n 2 his tells us that cos 2θ + i sin 2θ (cos θ + i sin θ) 2 cos 2 θ sin 2 θ + 2i cos θ sin θ. Comparing real and imaginary parts on left and right hand sides this gives you the double angle formulas cos θ cos 2 θ sin 2 θ and sin 2θ 2 sin θ cos θ. For n 3 you get, using the Binomial Theorem, or Pascal s triangle, so that and (cos θ + i sin θ) 3 cos 3 θ + 3i cos 2 θ sin θ + 3i 2 cos θ sin 2 θ + i 3 sin 3 θ cos 3 θ 3 cos θ sin 2 θ + i(3 cos 2 θ sin θ sin 3 θ) cos 3θ cos 3 θ 3 cos θ sin 2 θ sin 3θ cos 2 θ sin θ sin 3 θ. In this ay it is fairly easy to rite don similar formulas for sin 4θ, sin 5θ, etc.... 5. Calculus of complex valued functions A complex valued function on some interval I (a, b) R is a function f : I C. Such a function can be ritten as in terms of its real and imaginary parts, (8) f(x) u(x) + iv(x), in hich u, v : I R are to real valued functions.

7 One defines limits of complex valued functions in terms of limits of their real and imaginary parts. Thus e say that lim f(x) L x x 0 if f(x) u(x) + iv(x), L A + ib, and both lim u(x) A and x x 0 lim v(x) B x x 1 hold. From this definition one can prove that the usual limit theorems also apply to complex valued functions. Theorem 1. If lim x x0 f(x) L and lim x x0 g(x) M, then one has lim f(x) ± g(x) L ± M, x x 0 lim f(x)g(x) LM, x x 0 f(x) lim x x 0 g(x) L, provided M 0. M The derivative of a complex valued function f(x) u(x) + iv(x) is defined by simply differentiating its real and imaginary parts: (9) f (x) u (x) + iv (x). Again, one finds that the sum,product and quotient rules also hold for complex valued functions. Theorem 2. If f, g : I C are complex valued functions hich are differentiable at some x 0 I, then the functions f ± g, fg and f/g are differnetiable (assuming g(x 0 ) 0 in the case of the quotient.) One has (f ± g) (x 0 ) f (x 0 ) ± g (x 0 ) (fg) (x 0 ) f (x 0 )g(x 0 ) + f(x 0 )g (x 0 ) ( ) f (x 0 ) f (x 0 )g(x 0 ) f(x 0 )g (x 0 ) g g(x 0 ) 2 Note that the chain rule does not appear in this list! See problem 5 for more about the chain rule. 6. The Complex Exponential Function We finally give a definition of e a+bi. First e consider the case a 0, and e define (10) e it cos t + i sin t. for any real number t. Example. e πi cos π + i sin π 1. This leads to Euler s famous formula e πi + 1 0, hich combines the five most basic quantities in mathematics: e, π, i, 1, and 0. Reasons hy the definition (10) seems a good definition. Reason 1. We haven t defined e it before and e can do anything e like.

8 Reason 2. Substitute it in the Taylor series for e x : e it 1 + it + (it)2 2! + (it)3 3! + (it)4 4! + 1 + it t2 2! it3 3! + t4 4! + it5 5! 1 t 2 /2! + t 4 /4! + i ( t t 3 /3! + t 5 /5! ) cos t + i sin t. This is not a proof, because before e had only proved the convergence of the Taylor series for e x if x as a real number, and here e have pretended that the series is also good if you substitute x it. Reason 3. As a function of t the definition (10) gives us the correct derivative. Namely, using the chain rule (i.e. pretending it still applies for complex functions) e ould get de it ie it. dt Indeed, this is correct. To see this proceed from our definition (10): de it dt d cos t + i sin t dt d cos t + i d sin t dt dt sin t + i cos t i(cos t + i sin t) Reason 4. The formula e x e y e x+y still holds. Rather, e have e it+is e it e is. To check this replace the exponentials by their definition: e it e is (cos t + i sin t)(cos s + i sin s) cos(t + s) + i sin(t + s) e i(t+s). e a+bi in general. To define the complex exponential in general, e set e a+bi e a e ib e a (cos b + i sin b). One verifies as above in reason 3 that this gives us he right behaviour under differentiation. Thus, for any complex number r a + bi the function satisfies y(t) e rt e at (cos bt + i sin bt) y (t) dert re rt. dt This is all you need to kno to read the section in our text book on second order equations in the presence of imaginary roots. 7. Other handy things you can do ith complex numbers 7.1. Partial fractions. Consider the partial fraction decomposition x 2 + 3x 4 (x 2)(x 2 + 4) A x 2 + Bx + C x 2 + 4

9 The coefficent A is easy to find: multiply ith x 2 and set x 2 (or rather, take the limit x 2) to get A 22 + 3 2 4 2 2. + 4 Before e had no similar ay of finding B and C quickly, but no e can apply the same trick: multiply ith x 2 + 4, x 2 + 3x 4 (x 2) Bx + C + (x 2 A + 4) x 2, and substitute x 2i. This make x 2 + 4 0, ith result (2i) 2 + 3 2i 4 (2i 2) Simplify the complex number on the left: (2i) 2 + 3 2i 4 (2i 2) 2iB + C. 4 + 6i 4 2 + 2i 8 + 6i 2 + 2i ( 8 + 6i)( 2 2i) ( 2) 2 + 2 2 28 + 4i 8 7 2 + i 2 So e get 2iB + C 7 2 + i 2 ; since B and C are real numbers this implies B 7 4, C 1 2. 7.2. Complex amplitudes. A harmonic oscillation is given by y(t) A cos(ωt φ), here A is the amplitude, ω is the frequency, and φ is the phase of the oscillation. If you add to harmonic oscillations ith the same frequency ω, then you get another harmonic oscillation ith frequency ω. You can prove this using the addition formulas for cosines, but there s another ay using complex exponentials. It goes like this. Let y(t) A cos(ωt φ) and (t) B cos(ωt θ) be the to harmonic oscillations e ish to add. They are the real parts of Y (t) A{cos(ωt φ) + i sin(ωt φ) Ae iωt iφ Ae iφ e iωt Z(t) B{cos(ωt θ) + i sin(ωt θ) Be iωt iθ Be iθ e iωt Therefore y(t) + (t) is the real part of Y (t) + Z(t) hich is easy to compute: Y (t) + Z(t) Ae iφ e iωt + Be iθ e iωt ( Ae iφ + Be iθ) e iωt. If you no do the complex addition Ae iφ + Be iθ Ce iψ,

10 i.e. you add the numbers on the right, and compute the absolute value C and argument ψ of the sum, then e see that Y (t) + Z(t) Ce i(ωt ψ). Since e ere looking for the real part of Y (t) + Z(t), e get y(t) + (t) A cos(ωt φ) + B cos(ωt θ) C cos(ωt ψ). The complex numbers Ae iφ, Be iθ and Ce iψ are called the complex amplitudes for the harmonic oscillations y(t), (t) and y(t) + (t). The recipe for adding harmonic oscillations can therefore be summaried as follos: Add the complex amplitudes. 8. Problems (1) (a) Compute the folloing complex numbers by hand, and check the ansers on your calculator (if you don t have a calculator, check them ith someone else in your class). (b) Dra all numbers in the complex (or Argand ) plane (use graph paper or quad paper if necessary). (c) Compute absolute value and argument of all numbers involved. (1 + 2i)(2 i) (1 + i)(1 + 2i)(1 + 3i) (i.e. dra 1 + 2i, 2 i and the product; the same for the other problems) i 2 i 3 i 4 1 i ( 2 1 2 + i 2 2) 2 1 1 + i 5 ( 1 2 i 2 + i 2 3) 3 2e 3πi/4 e πi/3 e πi + 1 e 17πi/4 (2) Let θ, φ ( π 2, ) π 2 be to angles. (a) What are the arguments of 1 + i tan θ and of 1 + i tan φ? (Dra both and.) (b) Compute. (c) What is the argument of? (d) Compute tan(arg ). (3) Find formulas for cos 4θ, sin 4θ, cos 5θ and sin 6θ in terms of cos θ and sin θ, by using de Moivre s formula. 3 (4) In the folloing picture dra 2, 4, i, 2i, (2 + i) and (2 i). (Try to make a nice draing, use a ruler.) Make a ne copy of the picture, and dra, and. Make yet another copy of the draing. Dra 1/, 1/, and 1/. For this draing you need to kno here the unit circle is in your draing: Dra a circle centered at the origin of your choice, and let this be the unit circle.

+ c + di c d a + bi i(a + bi) b 90 2 2 3 a i bi a + bi ϕ 11 Figure 6. Picture for problem 4. (5) (a) Prove the folloing version of the Chain rule: If f : I C is a differentiable complex valued function, and g : J I is a differentiable real valued function, then h f g : J C is a differentiable function, and one has h (x) f (g(x))g (x). (b) Let n 0 be a nonnegative integer. Prove that if f : I C is a differentiable function, then g(x) f(x) n is also differentiable, and one has g (x) nf(x) n 1 f (x). Note that the chain rule from part (a) does not apply! Why? (6) Compute the derivatives of the folloing functions f(x) 1 x + i h(x) e ix2 g(x) log x + i arctan x k(x) log i + x i x Try to simplify your ansers. (7) (a) Verify directly from the definition that (b) Sho that (c) Sho that e it 1 e it. cos t eit + e it, sin t eit e it 2 2i cosh x cos ix, sinh x 1 sin ix i (8) Find A and φ here A cos(t φ) 12 cos(t π/6) + 8 cos(t π/4).

12 (9) The general solution of a second order linear differential equation contains expressions of the form Ae iβt +Be iβt. These can be reritten as C 1 cos βt+ C 2 sin βt. If Ae iβt + Be iβt 2 cos βt + 3 sin βt, then hat are A and B? (10) Compute (cos 2x ) 4 dx by using cos θ 1 2 (eiθ + e iθ ) and expanding the fourth poer. (11) Assuming a R, compute e 2x( sin ax ) 2 dx. (again by riting sin θ in terms of complex exponentials.)