We multiply complex numbers (a+bi)(c+di) just as we would the polynomials (a+bx)(c+dx) except that we remember that i 2 = 1.

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1 1 Complex Numbers 1.1 Introduction The complex number system is an extension of the real number system. It unifies the mathematical number system and explains many mathematical phenomena. We introduce a number i defined to satisfy the equation x 2 = 1. (As soon as we introduce this number, there is some ambiguity, for x = i also satisfies x 2 = 1!) The complex numbers are defined as all numbers of the form a + bi where a and b are real. We write C := {a + bi : a, b R}. A complex number of the form z = a + bi is said to have real part a and imaginary part b. Any number can be written in this form. The number i has real part 0 and is said to be purely imaginary ; the number 5 has imaginary part 0 and is real. The real numbers are a subset of the complex numbers. The conjugate of a complex number z = a + bi is z = a bi. Thus the conjugate of i is ī = i and the conjugate of 5 is merely Basic Operations We give the complex numbers a natural addition, subtraction and multiplication. We add and subtract complex numbers just as we would polynomials, keeping up with the real and imaginary parts. For example, ( + 4i) + (7 + 11i) = i and ( + 4i) (7 + 11i) = 4 7i. We multiply complex numbers (a+bi)(c+di) just as we would the polynomials (a+bx)(c+dx) except that we remember that i 2 = 1. For example, ( + 4x)(7 + 11x) = x + x + 44x 2 = x + 44x 2. So ( + 4i)(7 + 11i) = i + 44i 2 = i 44 = i. There is a natural understanding of division, but we want all our complex numbers to be written in Cartesian form a + bi so there is a little twist involved. Note that if z = a + bi then z z = (a + bi)(a bi) = a 2 + b 2. So if we are dividing by z, we may view 1 z as 1 z = z z z = a a 2 + b 2 b a 2 + b 2 i. Computationally, this means that anytime we have a fraction involving a complex number z in the denominator, we may multiply both numerator and denominator by z and simplify. For example, + 4i 7 11i = + 4i 7 11i i ( + 4i)(7 + 11i) i = = = i (7 11i)(7 + 11i) i

2 1. Geometric interpretation Mathematicians began to recognize the complex numbers sometime back in the Renaissance period (fifteenth and sixteenth centuries) but it was not until there was a geometric interpretation of the complex numbers that people began to feel comfortable with them. We may view the complex numbers as lying in the plane. Let the traditional x-axis represent the real numbers and the traditional y-axis represent the numbers of the form yi. We equate a complex number x + yi with the point (x, y). (So the imaginary numbers yi are at right angle to the real numbers!) For example, 2 + i can be graphed as the point (2, 1). The Cartesian plane (Argand diagram) In the Argand diagram, the complex number z = a+bi is equated with the point (a, b) in the Cartesian plane. So i = 0 + 1i is equated with the point (0, 1) and the number 1 = 1 + 0i is equated with the point (1, 0). The point (2, 1) represents the number 2 + i. The number ( + i)/2 is equated with the point ( /2, 1/2). The x-axis is called the real axis and the y-axis is called the imaginary axis. 1.4 Polar coordinates As soon as we have the geometry of the Cartesian plane at hand, we have a variety of powerful mathematical tools available. In particular, instead of using rectangular coordinates, we could use polar coordinates instead! Notice that the number ( + i)/2 is on the unit circle x 2 + y 2 = 1 which is at an angle of 0 degrees (= π/6) from the x-axis. In other words, ( + i)/2 = /2 + i/2 has polar coordinate form (r, θ) = (1, π/6). In general one can change a point (x, y) in the plane into polar coordinate form (r, θ) by noticing that (elementary trigonometry) and tan θ = y/x (1) r 2 = x 2 + y 2 (2) (the Pythagorean Theorem). One changes polar coordinates r and θ into Cartesian coordinates (x, y) by using the trigonometric definitions cos θ = x/r, sin θ = y/r, and so x = r cos θ () and x = r sin θ (4) 2

3 A point (x, y) can be written as (r cos θ, r sin θ) where r and θ are the polar coordinates of the point. In summary, x + iy = (x, y) = (r cos θ, r sin θ) = r cos θ + i r sin θ. For example, i = cos π 6 + i sin π 6. The value r represents the distance from the origin 0 to the point (x, y). This is called the modulus or absolute value, z, of the complex number z = x + iy. For example, the modulus of z = i = cos π 6 + i sin π is 1 since z is on the unit circle. 6 The modulus of z = + i is 2 since = 2 2. And the modulus of z = + 4i is 5 since = 5 2. Given any complex number z = x + iy we can change it into the polar coordinate form z = r(cos θ + i sin θ) which depends only on r and θ. Some call this the cis form (cos + i sin) form.) In this form i = cisπ 6 and z = + i = 2cis π 6. But there is an even better way to write complex numbers! That will be the subject of the next section.

4 Complex Numbers Worksheet 1 Basic Computations Perform the indicated operation. Simplify your answers so that the answers are in the form a + bi where a, b R. 1. (2 + i) + (4 + 5i). 2. (2 + i) + ( 2 + 5i).. (2 + i) + (4 i). 4. (2 + i) 5(4 + 5i). 5. 5(2 + i) 8(4 + 5i). 6. (2 + i)(4 + 5i). 7. (2 + i)(2 + i). 8. (2 + i)(2 i). 9. ( + 4i)( 4i). 10. (1 + i)(1 + i) i. 1 4i. 1 + i 1 2i. 2 + i 4 + 5i. 2 + i 2 i. 16. Write both square roots of -4 as complex numbers. 17. Find all solutions to x 2 + x + 1 = 0. Then use those solutions to factor x 2 + x Find all solutions to x 6 1 = 0. (Recall that A ± B factors as A ± B = (A ± B)(A 2 A + 1.) 19. Compute ( + 2i)(7 i). 20. Find the real and imaginary parts of ( + 2i)/(7 i). 21. Simplify 1/i. 4

5 Complex Numbers Worksheet 2 Polar Coordinates 1. Change the following rectangular coordinates into polar coordinates (a) (, 4) (b) (0, 1) (c) ( 1/2, /2) (d) (, 4) (e) (, 1) (f) (, 1) (g) ( 2, 2) (h) (5, 12) (i) (, ) (j) (1, 1) (k) (0, 0) 2. Change the following polar coordinates into rectangular coordinates (a) (r = 2, θ = π/4) (b) (5, tan 1 (/4)) (c) (2, π/6) (d) (1, π 4 ) (e) (5, tan 1 4 ) (f) (2, π 6 ) (g) (2, 7π 6 ) (h) (2, 601π 6 ) (i) ( 2, 601π 6 ) (j) ( 2 2, 7π 4 ). Give the Cartesian form (a + bi) of the following complex numbers. (a) cis( π 4 ) (b) 5 cis(tan 1 ( 4 )) (c) 2 cis( π 6 ) (d) 2 cis( 7π 6 ) (e) 2 cis( 601π 6 ) (f) 2 cis( 601π 6 ) (g) 2 2 cis( 7π 4 ) 5

6 1.5 Euler s formula and the right way to use complex numbers The value of complex numbers was recognized but poorly understood during the late Renaissance period. The number system was explicitly studied in the late 18th century. Euler used i for the square root of 1 in Gauss used the term complex in the early 1800 s. The Argand diagram or Gauss plane, was introduced in a memoir by Argand in Paris in 1806, although it was implicit in the doctoral dissertation of Gauss in 1799 and in work of Caspar Wessel around the same time. (See [Eves, Howard An Introduction to the History of Mathematics, 6th ed. Holt, Rinehart & Winston, New York, etc, 1990., page 479] for more historical notes.) Notice the following remarkable fact that if z 0 = i = cos π 6 + i sin π 6 then z 0 = i. (Multiply it out and see!) Thus z0 12 = 1 and so z 0 is a twelfth root of 1. Now the polar coordinate form for z 0 is r = 1, θ = π 6, that is, z 0 is exactly one-twelfth of the way around the unit circle. z 0 is a twelfth root of 1 and it is one-twelfth of the way around the unit circle. This is not a coincidence! DeMoivre apparently noticed this and proved (by induction) that if n is an integer then (cos θ + i sin θ) n = cos nθ + i sin nθ. (5) Thus exponentiation, that is raising a complex number to some power, is equivalent to multiplication of the arguments. Somehow the angles in the complex number act like exponents. Euler would explain why that was true. Let f(z) be a differentiable function and let f (n) (z) represent the n-th derivative of f(z) with respect to z. The Maclaurin polynomial (Taylor polynomial around zero) of degree n for f(z) is defined as f(0) + f (1) (0)z + f (2) (0) z2 2 + f () (0) z f (n 1) (0) zn 1 (n 1)! + f (n) (0) zn n!. Here is an exercise which follows work first done by Euler in the 1700 s. Let us find the Taylor polynomial of high degree (say of degree 9?) around zero for the following functions. 1. f(z) = exp(z) = e z 2. f(z) = cos z. f(z) = sin z 4. f(z) = exp(iz) = e iz. Using the work, above, write a formula for exp(iz) in terms of cosine and sine. We see, after this exercise, that the Taylor series of e iz is the same as the Taylor series of cos z +i sin z, that is, e iz = cos z + i sin z (6) If you recognize that (e iz ) n = e inz, then you have the DeMoivre formula in (5). This explains the coincidence above. If z = cos π 6 + i sin π 6 is one-twelfth of the way around the unit circle then raising z to the twelfth power will simply multiply the angle θ by twelve and move the point to the point with angle zero: (1, 0) = 1 + 0i. We may use Euler s formula and write a formula for cos z as a sum of exponential functions. Since e iz = cos(z) + i sin(z) then e iz = cos( z) + i sin( z) = cos(z) i sin(z). (7) 6

7 Add the expressions for e iz and e iz to get and so e iz + e iz = 2 cos(z) (8) cos z = eiz + e iz. (9) 2 If we subtract the equation e iz = cos z i sin z from Euler s equation (6) and then divide by 2i, we have a formula for sine: sin z = eiz e iz. (10) 2i Euler s formula writes the exponential function in terms of trig functions; now we have written a trig function in terms of exponential functions. The exponential and trig functions are very closely related. Trig functions are, in some sense, really exponential functions in disguise! Complex numbers v. Real numbers Here are some things one can do with the real numbers: 1. Find an infinite set of numbers, x, such that sin(x) = 1/2. 2. Find a number x such that e x = Compute ln(2). Here are some things that require complex numbers: 1. Show that f(x) = e x is periodic with period 2πi, that is, f(x + 2πi) = f(x). 2. Find an infinite set of numbers, x, such that e x = 1/2.. Use Euler s formula to create the formula for sin z as a sum of exponential functions. 4. Find a number x such that sin(x) = Compute ln( 2). 7

8 Complex Numbers Worksheet Euler s Formula 1. Put the following complex numbers into polar coordinates re iθ Give two different answers, using different values of θ. (a) + 4i (b) + i (c) + i (d) 2 2i (e) i (f) i (g) 1 i (h) 0 2. Write the polar form re iθ for the following complex numbers. (a) 1. (b) 2i. (c) 2 + i 2. (d) 2 i 2. (e) i. (f) + i.. Write the polar form re iθ for the following complex numbers. Then give your answer in Cartesian form z = x + iy. (a) (1 + i ). (b) ( 2 i 2) 6. (c) ( 2 + i 2) 84 (d) ( 2 i 2) 24 (e) ( i) 00 (f) ( + i) Compute ( i) Write the polar form re iθ for the following complex numbers. Then give your answer in Cartesian form z = x + iy. (a) (8 + 8i) 10 (b) (2 + 2i) Write the polar form re iθ for the following complex numbers. Then give your answer in Cartesian form z = x + iy. (a) ( 1 + i). (b) ( 1 + i) 20. (c) ( 1 + i) 1/. (d) i. 8