Complex Numbers James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 19, 2013 Outline Complex Numbers Complex Number Calculations Complex Functions Calculations With Complex Functions
Abstract This lecture is going discuss complex numbers. When we use the quadratic equation to find the roots of a polynomial like f (t) = t 2 + t + 1, we find t = 1 ± 1 4 2 = 1 2 ± 3 2 Since, it is well known that there are no numbers in our world whose squares can be negative, it was easy to think that the term 3 represented some sort of imaginary quantity. But, it seemed reasonable that the usual properties of the function should hold. Thus, we can write 3 = 1 3
The term 1 had the amazing property that when you squared it, you got back 1! Thus, the square root of any negative number c for a positive c could be rewritten as 1 c. It became clear to the people studying the roots of polynomials such as our simple one above, that if the set of real numbers was augmented to include numbers of the form ± 1 c, there would be a nice way to represent any root of a polynomial. Since c can be any positive number, it seemed like two copies of the real numbers were needed: one that was the usual real numbers and another copy which was any real number times this strange quantity 1. It became very convenient to label the set of all traditional real numbers as the x axis and the set of numbers prefixed by 1 as the y axis. Since the prefix 1 was the only real difference between the usual real numbers and the new numbers with the prefix 1, this prefix 1 seemed the the quintessential representative of this difference. Historically, since these new prefixed numbers were already thought of as imaginary, it was decided to start labeling 1 as the simple letter i where i is short for imaginary!
Thus, a number of this sort could be represented as a + b i where a and b are any ordinary real numbers. In particular, the roots of our polynomial could be written as t = 1 2 ± i 3 2. A generic complex number z = a + b i can thus be graphed as a point in the plane which has as its horizontal axis the usual x axis and as its vertical axis the new i y axis. We call this coordinate system the Complex Plane. The magnitude of the complex number z is defined to be the length of the vector which connects the ordered pair (a, b) in this plane with the origin (0, 0). This is labeled as r. This magnitude is also called the modulus or magnitude of z and is represented by the same symbol we use for the absolute value of a number, z. Here this magnitude is the length of the hypotenuse of the triangle consisting of the points (0, 0), (a, 0) and (a, b) in the complex plane. Hence, z = (a) 2 + (b) 2
The angle measured from the positive x axis to this vector is angle associated with the complex number z and is denoted by the symbol θ or Arg(z). Hence, there are two equivalent ways we can represent a complex number z. 1. We can use coordinate information and write z = a + b i 2. We can use magnitude and angle information. Look at the figure below and you can see a = r cos(θ) and b = r sin(θ). Thus, z = z (cos(θ) + i sin(θ)) = r (cos(θ) + i sin(θ)) The complex conjugate of z = a + b i is called z and is z = a b i. i y b r z = a + bi θ a x A complex number a + b i has real part a and imaginary part b. The coordinate (a, b) is graphed in the usual Cartesian manner as an ordered pair in the x + iy complex plane. The magnitude of z is (a) 2 + (b) 2 which is shown on the graph as r. The angle associated with z is drawn as an arc of angle θ
i y -b z = a bi a x θ r A complex number z = a + b i has real part a and imaginary part b. Its complex conjugate is z = a b i. The coordinate (a, b) is graphed in the usual Cartesian manner as an ordered pair in the complex plane. The magnitude of z is (a) 2 + (b) 2 which is shown on the graph as r. The angle associated with z is drawn as an arc of angle θ or 2π θ. So if z = a + b i, we also know z = r (cos(θ) + i sin(θ)). The angle for z is then θ. Thus ( ) z = r cos( θ) + i sin( θ). But cos is even (i.e. cos( x) = cos(x) and sin is odd (i.e. sin( x) = sin(x). So ( ) z = r cos(θ) i sin(θ).
Example For the complex number z = 2 + 4 i 1. Find its magnitude 2. Write it in the form r (cos(θ) + i sin(θ)) using radians 3. Graph it in the complex plane showing angle in degrees Solution The complex number 2 + 4 i has magnitude (2) 2 + (4) 2 which is 20. Since both the real and imaginary component of the complex number are positive, this complex number is graphed in the first quadrant of the complex plane. Hence, the angle associated with z should be between 0 and π/2. We can easily calculate the angle θ associated with z to be tan 1 (4/2). Example For the complex number 2 + 8 i 1. Find its magnitude 2. Write it in the form r (cos(θ) + i sin(θ)) using radians 3. Graph it in the complex plane showing angle in degrees Solution 2 + 8 i has magnitude ( 2) 2 + (8) 2 = 68. This complex number is graphed in the second quadrant of the complex plane because the real part is negative and the imaginary part is positive. Hence, the angle associated with z should be between π/2 and π. The angle θ is then π tan 1 ( 8/ 2 ) Hence, θ is 1.82 radians or 104.04 degrees. The graph in the next figure is instructive.
Im(z) Re(z) r i y z = 2 + 8i θ x A complex number 2 + 8 i has real part 2 and imaginary part 8. The coordinate ( 2, 8) is graphed in the usual Cartesian manner as an ordered pair in the complex plane. The magnitude of z is ( 2) 2 + (8) 2 which is shown on the graph as r. The angle associated with z is drawn as an arc of angle θ Homework 29 For each complex number, find its magnitude, its associated angle, write it in the form r (cos(θ) + i sin(θ)), find its complex conjugate and graph it carefully. 29.1 z = 3 + 6 i. 29.2 z = 3 6 i. 29.3 z = 3 2 i. 29.4 z = 5 + 3 i. 29.5 z = 4 3 i. 29.6 z = 5 7 i.
Let z = r (cos(θ) + i sin(θ)) and let s think of θ are a variable now. So we have defined a function f (θ) = r (cos(θ) + i sin(θ)). Let s find f (θ). We have f (θ) = r sin(θ) + i r cos(θ) = i 2 r sin(θ) + i r cos(θ) ( ) = i ri sin(θ) + r cos(θ) = i f (θ). So f (θ) = i f (θ) or f (θ) f (θ) = i. Taking the antiderivative of both sides, this suggests ln(f (θ)) = i f (θ) = e iθ. Our antiderivative argument suggested we define e i θ = r (cos(θ) + i sin(θ)) and by direct calculation, we had (e i θ ) = i e i θ. We can also show that the usual laws for exponential functions will hold. e a+bi θ = e aθ e i bθ And the usual product rule and chain rule apply to derivatives. ( e (a +b i) θ ) = ( ) ) ) e aθ e i bθ = (e aθ e i bθ + e (e aθ i bθ = ae aθ e i bθ + i be aθ e i bθ = (a + ib)e (a +b i)θ.
Example For the complex function e ( 2 +8 i)t 1. Find its magnitude 2. Write it in its fully expanded form 3. Find its derivative Solution We know that exp (( 2 + 8 i) t) = e 2t e 8it = e 2t (cos(8t) + i sin(8t)). Since the complex magnitude of e 8it is always one, we see e ( 2 +8 i)t = e 2t e 8it = e 2t The derivative is ( 2 + 8 i) e ( 2 +8 i)t. Example For the complex function e ( 1 +2 i)t 1. Find its magnitude 2. Write it in its fully expanded form 3. Find its derivative. Solution We have exp (( 1 + 2 i) t) = e t e 2it = e t (cos(2t) + i sin(2t)). Since the complex magnitude of e 2it is always one, we see e ( 1 +2 i)t = e t The derivative is ( 1 + 2 i) e ( 1 +2 i)t.
Example For the complex function e 2i t 1. Find its magnitude 2. Write it in its fully expanded form 3. Find its derivative. Solution We have exp ((0 + 2 i) t) = e 0t e 2it = cos(2t) + i sin(2t). Since the complex magnitude of e 2it is always one, we see e (0 +2 i)t = 1 The derivative is (0 + 2 i) e (0 +2 i)t. Homework 30 For each complex function, find its magnitudes, write it in expanded form and find its derivative. 30.1 e ( 2 +3 i) t. 30.2 e ( 1 +4 i) t. 30.3 e (0 +14 i) t. 30.4 e ( 8 9 i) t. 30.5 e ( 2 +π i) t.