Complex Numbers and Exponentials Definition and Basic Operations A complexnumber is nothing morethan a point in the xy plane. The first component, x, ofthe complex number (x,y) is called its real part and the second component, y, is called its imaginary part, even though there is nothing imaginary (1) about it. The sum and product of two complex numbers (x 1,y 1 ) and (x,y ) are defined by (x 1,y 1 )+(x,y ) = (x 1 +x,y 1 +y ) (x 1,y 1 )(x,y ) = (x 1 x y 1 y,x 1 y +x y 1 ) respectively. We ll get an effective memory aid for the definition of multiplication shortly. It is conventional to use the notation x+iy (or in electrical engineering x+jy) to stand for the complex number (x,y). In other words, it is conventional to write x in place of (x,0) and i in place of (0,1). In this notation, the sum and product of two complex numbers z 1 = x 1 +iy 1 and z = x +iy is given by z 1 +z = (x 1 +x )+i(y 1 +y ) z 1 z = x 1 x y 1 y +i(x 1 y +x y 1 ) Addition and multiplication of complex numbers obey the familiar algebraic rules z 1 +z = z +z 1 z 1 z = z z 1 z 1 +(z +z 3 ) = (z 1 +z )+z 3 z 1 (z z 3 ) = (z 1 z )z 3 0+z 1 = z 1 1z 1 = z 1 z 1 (z +z 3 ) = z 1 z +z 1 z 3 (z 1 +z )z 3 = z 1 z 3 +z z 3 The negative of any complex number z = x+iy is defined by z = x+( y)i, and obeys z +( z) = 0. The inverse, z 1 or 1 z, of any complex number z = x+iy, other than 0, is defined by 1 zz = 1. We shall see below that it is given by the formula 1 z = x x +y + y x +y i. The complex number i has the special property i = (0+1i)(0+1i) = (0 0 1 1)+i(0 1+1 0) = 1 To remember how to multiply complex numbers, you just have to supplement the familiar rules of the real number system with i = 1. For example, if z = 1+i and w = 3+4i, then z +w = (1+i)+(3+4i) = 4+6i zw = (1+i)(3+4i) = 3+4i+6i+8i = 3+4i+6i 8 = 5+10i (1) Don t attempt to attribute any special significance to the word complex in complex number, or to the word real in real number and real part, or to the word imaginary in imaginary part. All are just names. c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 1
Other Operations The complex conjugate of z is denoted z and is defined to be z = x iy. That is, to take the complex conjugate, one replaces every i by i. Note that z z = (x+iy)(x iy) = x ixy +ixy +y = x +y is always a positive real number. In fact, it is the square of the distance from x+iy (recall that this is the point (x,y) in the xy plane) to 0 (which is the point (0,0)). The distance from z = x+iy to 0 is denoted z and is called the absolute value, or modulus, of z. It is given by z = x +y = z z Since z 1 z = (x 1 +iy 1 )(x +iy ) = (x 1 x y 1 y )+i(x 1 y +x y 1 ), z 1 z = (x 1 x y 1 y ) +(x 1 y +x y 1 ) = x 1 x x 1x y 1 y +y1 y +x 1 y +x 1y x y 1 +x y 1 = x 1 x +y1 y +x1 y +x y1 = (x 1 +y 1 )(x +y ) = z 1 z for all complex numbers z 1,z. Since z = z z, we have z ( ) z z = 1 for all complex numbers z 0. This says that the multiplicative inverse, 1 z, of any nonzero complex number z = x+iy is z 1 = z z = x iy x +y = x x +y y x +y i This is the formula for 1 z given above. It is easy to divide a complex number by a real number. For example 11+i 5 = 11 5 + 5 i In general, there is a trick for rewriting any ratio of complex numbers as a ratio with a real denominator. For example, suppose that we want to find 1+i 3 4i 3+4i. The trick is to multiply by 1 = 3 4i. The number 3 4i is the complex conjugate of the denominator 3+4i. Since (3+4i)(3 4i) = 9 1i+1i+16 = 5 1+i 3+4i = 1+i 3 4i 3+4i 3 4i = (1+i)(3 4i) 5 = 11+i 5 = 11 5 + 5 i ThenotationsRez andimz standfortherealandimaginarypartsofthecomplexnumberz,respectively. If z = x+iy (with x and y real) they are defined by Rez = x Imz = y Note that both Rez and Imz are real numbers. Just subbing in z = x iy gives Rez = 1 (z + z) Imz = 1 i (z z) c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials
The Complex Exponential Definition and Basic Properties. For any complex number z = x+iy the exponential e z, is defined by e x+iy = e x cosy +ie x siny In particular, e iy = cosy+isiny. This definition is not as mysterious as it looks. We could also define e iy by replacing x with iy in the Taylor series expansion e x = n=0 xn n!. The even terms in this expansion are e iy = 1+iy + (iy)! + (iy)3 3! + (iy)4 4! + (iy)5 5! + (iy)6 6! + 1+ (iy)! + (iy)4 4! + (iy)6 6! + = 1 y! + y4 4! y6 6! + = cosy and the odd terms in this expansion are iy+ (iy)3 3! + (iy)5 5! + = i (y y3 For any two complex numbers z 1 and z 3! + y5 e z1 e z = e x1 (cosy 1 +isiny 1 ) e x (cosy +isiny ) = e x1+x (cosy 1 +isiny 1 )(cosy +isiny ) 5! + ) = isiny = e x1+x {(cosy 1 cosy siny 1 siny )+i(cosy 1 siny +siny 1 cosy )} = e x1+x {cos(y 1 +y )+isin(y 1 +y )} = e (x1+x)+i(y1+y) = e z1+z so that the familiar multiplication formula for real exponentials also applies to complex exponentials. For any complex number c = α+iβ and real number t so that the derivative with respect to t is also the familiar one. e ct = e αt+iβt = e αt [cos(βt)+isin(βt)] d dt ect = αe αt [cos(βt)+isin(βt)]+e αt [ βsin(βt)+iβcos(βt)] = (α+iβ)e αt [cos(βt)+isin(βt)] = ce ct Relationship with sin and cos. When θ is a real number e iθ = cosθ +isinθ e iθ = cosθ isinθ = e iθ (1) are complex numbers of modulus one. Solving for cosθ and sinθ (by adding and subtracting the two equations) cosθ = 1 (eiθ +e iθ ) = Ree iθ sinθ = 1 i (eiθ e iθ ) = Ime iθ c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 3
These formulae make it easy derive trig identities. For example cosθcosφ = 1 4 (eiθ +e iθ )(e iφ +e iφ ) = 1 4 (ei(θ+φ) +e i(θ φ) +e i( θ+φ) +e i(θ+φ) ) = 1 4 (ei(θ+φ) +e i(θ+φ) +e i(θ φ) +e i( θ+φ) ) [ ] cos(θ +φ)+cos(θ φ) = 1 and, using (a+b) 3 = a 3 +3a b+3ab +b 3, sin 3 ( θ = 1 8i e iθ e iθ) 3 ( = 1 8i e i3θ 3e iθ +3e iθ e i3θ) = 3 1 4 i( e iθ e iθ) ( 1 1 4 i e i3θ e i3θ) = 3 4 sinθ 1 4 sin(3θ) and cos(θ) = Ree θi = Re ( e iθ) = Re ( cosθ+isinθ ) = Re ( cos θ+isinθcosθ sin θ ) = cos θ sin θ Polar Coordinates. Let z = x + iy be any complex number. Writing (x,y) in polar coordinates in the usual way gives x = rcosθ, y = rsinθ and y x+iy = re iθ x+iy = rcosθ +irsinθ = re iθ r In particular θ x ( 1,0)= 1 y i=(0,1) π π 1=(1,0) π i=(0, 1) x 1 = e i0 = e πi = e kπi for k = 0,±1,±, 1 = e iπ = e 3πi = e (1+k)πi for k = 0,±1,±, i = e iπ/ = e 5 πi = e (1 +k)πi for k = 0,±1,±, i = e iπ/ = e 3 πi = e ( 1 +k)πi for k = 0,±1,±, The polar coordinate θ = tan 1 y x associated with the complex number z = x+iy, i.e. the point (x,y) in the xy plane, is also called the argument of z. The polar coordinate representation makes it easy to find square roots, third roots and so on. Fix any positive integer n. The n th roots of unity are, by definition, all solutions z of z n = 1 c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 4
Writing z = re iθ r n e nθi = 1e 0i The polar coordinates (r,θ) and (r,θ ) represent the same point in the xy plane if and only if r = r and θ = θ +kπ for some integer k. So z n = 1 if and only if r n = 1, i.e. r = 1, and nθ = kπ for some integer k. The n th roots of unity are all the complex numbers e πi k n with k integer. There are precisely n distinct n th roots of unity because e πi k n = e πik n if and only if π k n πk n = πk k n is an integer multiple of π. That is, if and only if k k is an integer multiple of n. The n distinct nth roots of unity are y e πi 6 e πi 1 6 1, e πi 1 n, e πi n, e πi 3 n,, e πi n 1 n e πi3 6= 1 1=e πi0 6 x e πi4 6 e πi 5 6 Exploiting Complex Exponentials in Calculus Computations You have learned how to evaluate integrals involving trigonometric functions by using integration by parts, various trigonometric identities and various substitutions. It is often much easier to just use (1). Here are two examples Example 1 [e (1+i)x +e (1 i)x] dx e x cosx dx = 1 e x[ e ix +e ix] dx = 1 = 1 [ 1 1+i e(1+i)x + 1 1 i e(1 i)x] +C This form of the indefinite integral looks a little weird because of the i s. But it is correct and it is purely real, despite the i s, because 1 1 i e(1 i)x 1 is the complex conjugate of 1+i e(1+i)x. We can convert the indefinite integral into a more familiar form just by subbing back in e ±ix 1 = cosx±isinx, 1+i = 1 i (1+i)(1 i) = 1 i and 1 1 i = 1 1+i = 1+i. e x cosx dx = 1 ex[ 1 1+i eix + 1 1 i e ix] +C = 1 ex[ 1 i 1+i (cosx+isinx)+ (cosx isinx)] +C = 1 ex cosx+ 1 ex sinx+c Example Using (a+b) 4 = a 4 +4a 3 b+6a b +4ab 3 +b 4, [e cos 4 x dx = 1 ix +e ix] 4 dx = 1 [e 4ix 4 +4e ix +6+4e ix +e 4ix] dx 4 = 1 4 [ 1 4i e4ix + 4 i eix +6x+ 4 i e ix + 1 [ = 1 1 1 4 4i e 4ix] +C i (e4ix e 4ix )+ 4 i (eix e ix )+6x ] +C = 1 4 [ 1 sin4x+4sinx+6x] +C = 1 3 sin4x+ 1 4 sinx+ 3 8 x+c c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 5
Complex exponentials are also widely used to simplify the process of guessing solutions to ordinary differential equations. Here are two examples. Example 3 The general solution to the ordinary differential equation y (t)+4y (t)+5y(t) = 0 () is of the form C 1 u 1 (t)+c u (t) with u 1 (t) and u (t) being two (independent) solutions to () and with C 1 and C being arbitrary constants. The easiest way to find u 1 (t) and u (t) is to guess them. And the easiest way to guess them is to try () y(t) = e rt, with r being a constant to be determined. Substituting y(t) = e rt into () gives r e rt +4re rt +5e rt = 0 ( r +4r+5 ) e rt = 0 r +4r+5 = 0 This quadratic equation for r can be solved either by using the high school formula or by rewriting it r +4r+5 = 0 (r+) +1 = 0 (r +) = 1 r+ = ±i r = ±i So the general solution to () is y(t) = C 1 e ( +i)t +C e ( i)t This is one way to write the general solution, but there are many others. In particular there are quite a few people in the world who are (foolishly) afraid of complex exponentials. We can hide them by using (1). y(t) = C 1 e ( +i)t +C e ( i)t = C 1 e t e it +C e t e it = C 1 e t( cost+isint ) +C e t( cost isint ) = (C 1 +C )e t cost+(ic 1 ic )e t sint = D 1 e t cost+d e t sint with D 1 = C 1 + C and D = ic 1 ic being two other arbitrary constants. Don t make the mistake of thinking that D must be complex because i appears in the formula D = ic 1 ic relating D and C 1,C. Noone said that C 1 and C are real numbers. In fact, in typical applications, the arbitrary constants are determined by initial conditions and often D 1 and D turn out to be real and C 1 and C turn out to be complex. For example, the initial conditions y(0) = 0, y (0) = force 0 = y(0) = C 1 +C = y (0) = ( +i)c 1 +( i)c The first equation gives C = C 1 and then the second equation gives ( +i)c 1 ( i)c 1 = ic 1 = ic 1 = 1 C 1 = i, C = i and D 1 = C 1 +C = 0 D = ic 1 ic = () The reason that y(t) = e rt is a good guess is that, with this guess, all of y(t), y (t) and y (t) are constants times e rt. So the left hand side of the differential equation is also a constant, that depends on r, times e rt. So we just have to choose r so that the constant is zero. c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 6
Example 4 We shall now guess one solution (i.e. a particular solution) to the differential equation y (t)+y (t)+3y(t) = cost (3) Equations like this arise, for example, in the study of the RLC circuit. We shall simplify the computation by exploiting that cost = Ree it. First, we shall guess a function Y(t) obeying Y +Y +3Y = e it (4) Then, taking complex conjugates, Ȳ +Ȳ +3Ȳ = e it ( 4) and, adding 1 () and 1 ( ) together will give (ReY) +(ReY) +3(ReY) = Ree it = cost which shows that ReY(t) is a solution to (1). Let s try Y(t) = Ae it. This is a solution of () if and only if ( d dt Ae it ) ( + d dt Ae it ) +3Ae it = e it (+i)ae it = e it So we have found a solution to () and Re eit +i coordinates. So A = 1 +i is a solution to (1). To simplify this, write +i in polar +i = e iπ 4 e it +i = eit e i π 4 = 1 ei(t π 4 ) Re eit +i = 1 cos(t π 4 ) c Joel Feldman. 014. All rights reserved. August 6, 014 Complex Numbers and Exponentials 7