Complex Numbers. Integers, Rationals, and Reals. The natural numbers: The integers:

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Complex Numbers Integers, Rationals, and Reals The natural numbers: N {... 3, 2,, 0,, 2, 3...} The integers: Z {... 3, 2,, 0,, 2, 3...} Note that any two integers added, subtracted, or multiplied together give another integer. We cannot necessarily divide, though. I can tell you that multiplying 3 by 2 gets you 6, and thus that dividing 6 by 2 gets you 3; some division problems work out. On the other hand, others don t; you can t find an integer that tells you what happens when you divide 5 by 2. The rational numbers are the set of all fractions of integers: So Q { p }, p, q Z q Natural numbers let us count, add, and multiply. Integers let us subtract. Rational numbers let us divide.

There all still things we would like to do that we cannot do; the simplest of these is solving polynomial equations. Again, just like the rationals allowed some division, they will also allow some equations to be solved. If I want an x so that x 2 2 things get trickier. Much trickier, as no rational number will do the trick. 2

So 2 is irrational, and we make the real numbers to include it and other things we need to solve polynomial equations. There are any number of definitions, but here are two somewhat intuitive ones:. The rational numbers are the set of all numbers made from whole numbers and decimals that either truncate or repeat. The reals are all decimals, period: all the rationals, plus any decimal that doesn t truncate or repeat, too. 2. The real numbers are all the stuff in between the rationals. For example, while we can t have a 2 that is rational, we can have rational numbers that are arbitrarily close, above and below. So 2 [, 2] [ 7 5, 0 ] 7 [ 393 985, 577 ] 408 [ 4732 3346, 4243 ] 80782 Complex Numbers We immediately find polynomials that we still cannot solve within the reals; for instance, x 2 + 0 This gives us x ± which cannot be approximated by nearby rationals or expressed as a repeating or non-repeating decimal. And, just as we must use different symbols for irrational numbers ( 2, π, etc.) we now need a new symbol: i i is imaginary, and the extension of the reals by i gives us the complex numbers. Complex numbers have real and imaginary parts; if c is a complex number, then c a + ib a, b R This is it we re done. We can (as we ll see in a minute) add, subtract, multiply, and divide, and all polynomials even ones with complex coefficients have solutions. 3

Note that all imaginary numbers are complex (with a 0) and all real numbers are complex (with b 0). We call a and b the real and imaginary parts of a complex number, with Re(c) a Im(c) b Arithmetic with complex numbers is straightforward, but you have to keep your is separate. Adding two complex numbers gives c a + ib c 2 a 2 + ib 2 c + c 2 (a + a 2 ) + i(b + b 2 ) and subtracting c c 2 (a a 2 ) + i(b b 2 ) Multiplying a complex number by a real number involves distributing the real number c a + ib kc k(a + ib) ka + i(kb) Multiplying two complex numbers requires FOILing out the expression like it was a polynomial: c c 2 (a + ib ) (a 2 + ib 2 ) a a 2 + a (ib 2 ) + ib (a 2 ) + (ib )(ib 2 ) a a 2 + i(a b 2 + b a 2 ) + i 2 b b 2 a a 2 + i(a b 2 + b a 2 ) + ( )b b 2 (a a 2 b b 2 ) + i(a b 2 + b a 2 ) 4

The Complex Plane and Complex Conjugation Once we have a complex number c, we can figure out a and b so that c a + ib and, conversely, if we have a and b, we get a unique c. We can associate each complex number with a pair of real numbers, then: c (a, b) We can look at these as coordinates, and draw this out with a Cartesian coordinate system. By convention, the left-right (x) axis is the real axis, and the updown (y) axis imaginary. Every complex number is a point on this plane. Alternately, we can look at the (a, b) pair as a twodimensional vector. Adding two complex numbers is the same as adding their real and imaginary components, separately, so complex-number addition is the same as vector addition 5

Multiplying by a real number is the same as scaling the size of the vector right on through the origin if the real number is negative: Interestingly, multiplying a complex number by i is the same as rotating it 90 counter-clockwise. A particularly interesting operation is reflection across the x axis: This operation takes a + ib to a ib, and is called complex conjugation, denoted by an asterisk. We write (a + ib) a ib Note that any real number has a complex conjugate equal to itself a a and that the complex conjugate of the complex conjugate is just the original number: (c ) c If we like, we can derive Re(c) 2 (c + c ) Im(c) 2i (c c ) [(a + ib) (a ib)] 2i (a + ib a + ib) 2i 2i (2ib) b Multiplication: cc (a + ib)(a ib) a 2 a(ib) + ib(a) i 2 b 2 6

a 2 ( )b 2 a 2 + b 2 We define a number times its complex conjugate as the square modulus c 2 cc and if a and b are real and they always are then The square modulus of a complex number c is always a positive real number, and it is only equal to 0 if c 0. The modulus or absolute value of a complex number is just c c 2 a 2 + b 2 Geometrically, this is just the distance of point c to the origin in the complex plane, or it is the length of the vector represented by c. The complex conjugate and the square modulus give us a good way to divide complex numbers: c a + ib c a + ib c a ib cc (a + ib)(a ib) c a ib c 2 a 2 i 2 b 2 a ib a 2 + b 2 a a 2 + b 2 i b a 2 + b 2 and once we do the above we can multiply to figure out c /c 2 for any two complex numbers. Polar Form of Complex Numbers Every complex number has a modulus, which is its distance to the origin, and a direction, which tells whether it is positive, negative, 7

i-ive, or i-ive... or somewhere in between. If we put this direction in terms of the angle to the real number line, we get c a + ib a c cos θ b c sin θ If we go ahead an call the modulus r, we just have the transformation to polar coordinates: defined by (a, b) (r, θ) a r cos θ b r sin θ c a + ib r cos θ + ir sin θ r(cos θ + i sin θ) If we like, we can also get r a 2 + b 2 tan θ y x The Euler Relation Taylor series refresher: cos x 2 x2 + 4! x4 6! x6 +... ( ) k (2k)! x2k k0 sin x x 3! x3 + 5! x5 7! x7 +... ( ) k (2k + )! x2k+ k0 e x + x + 2 x2 + 3! x3 +... k0 x k k! Let s rewrite these a little, using i 2, i 3 i, and i 4. We get cos x 2 x2 + 4! x4 6! x6 +... 8

+ i 2 2 c2 + i 4 4! x4 + i 6 6! x6 +... + 2 (ix)2 + 4! (ix)4 + 6! (ix)6 +... n0,2,4,... n! (ix)n Similarly sin x x 3! x3 + 5! x5 7! x7 +... x + i 2 3! x3 + i 4 5! x5 + i 6 7! x7 +... i sin x ix + 3! (ix)3 +... n,3,5,... n! (ix)n Putting these together gives us Euler s relation: cos x + i sin x + ix + 2 (ix)2 + 3! (ix)3 +... n0 e ix n! (ix)n 9

We can use Euler s relation to express the sine and cosine in terms of complex exponentials: e iθ cos θ + i sin θ e iθ e i( θ) cos( θ) + i sin( θ) cos θ i sin θ e iθ + e iθ cos θ + i sin θ + cos θ i sin θ 2 cos θ cos θ ( e iθ + e iθ) 2 e iθ e iθ cos θ + i sin θ (cos θ i sin θ) 2i sin θ sin θ ( e iθ e iθ) 2i You can use these to avoid ever memorizing another trigonometric identity. sin 2 θ (sin θ) 2 [ ( e iθ e iθ)] 2 2i ( e iθ 4i 2 e iθ e iθ e iθ e iθ e iθ + e iθ e iθ) ( e 2iθ e 0 e 0 + e 2iθ) 4 ( e 2iθ + e 2iθ 2 ) 4 (2 cos 2θ 2) 4 2 cos 2θ 2 and to simplify integrals that would otherwise look impossible [ sin t e αt ( dt e it e it)] e αt dt 0 0 2i ( e it αt e it αt) dt 2i 0 ( e (α i)t e (α+i)t) dt 2i 0 [ ] e (α i)t [ ] e (α+i)t 2i (α i) 0 2i (α + i) 0 [ ] 0 [ ] 0 2i (α i) 2i (α + i) ( 2i α i ) α + i 0

[ ] (α + i) 2i (α i)(α + i) (α i) (α i)(α + i) ( ) α + i α + i 2i α 2 i 2 2i 2i α 2 + α 2 + de Moivre s Formula It s important to note that we can use all of the properties of exponential functions with complex numbers. So e inθ e inθ e i(nθ) e (iθ)n cos nθ + i sin nθ (e iθ ) n cos nθ + i sin nθ (cos θ + i sin θ) n This is de Moivre s Formula. We can also figure out sines and cosines of imaginary (or complex) arguments. cos ix [e i(ix) + e i(ix)] 2 ( e x + e x) 2 cosh x Similarly, sin ix i sinh x