Quick Overview Complex Numbers 2012

Quick Overview: Complex Numbers February 24, 2012 1 Initial Definitions Definition 1 The complex number z is defined as: z = a + bi where a, b are real numbers and i = √ (1) −1. Remarks about the definition: • Engineers typically use j instead of i. • Examples of complex numbers: 5 + 2i, • Powers of i: 3− √ 2i, i2 = −1 i4 = 1 i6 = −1 3, −5i i3 = −i i5 = i i7 = −i .. . • All real numbers are also complex (by taking b = 0). 2 Visualizing Complex Numbers A complex number is defined by it’s two real numbers. If we have z = a + bi, then: Definition 2 The real part of a + bi is a, Re(z) = Re(a + bi) = a The imaginary part of a + bi is b, Im(z) = Im(a + bi) = b To visualize a complex number, we use the complex plane C, where the horizontal (or x-) axis is for the real part, and the vertical axis is for the imaginary part. That is, a + bi is plotted as the point (a, b). In Figure 1, we can see that it is also possible to represent the point a + bi, or (a, b) in polar form, by computing its modulus (or size), and angle (or argument): p θ = arg(z) r = |z| = a2 + b2 We have to be a bit careful defining φ, since there are many ways to write φ (and we could add multiples of 2π as well). Typically, the argument of the complex number z = a + bi is defined to be the 4-quadrant “inverse tangent”1 that returns −π < θ ≤ π. 1 For example, in Maple this special angle is computed as arctan(b,a), and in Matlab the command is atan2(b,a). 1 Im(z) 4i 3i z = a + bi 2i r b 1i θ Re(z) a −1i Figure 1: Visualizing z = a + bi in the complex plane. Shown are the modulus (or length) r and the argument (or angle) θ. That is, formally we can define         θ = arg(a + bi) =        the argument as: tan−1 (b/a) tan−1 (b/a) + π tan−1 (b/a) − π π/2 −π/2 Undefined if if if if if if a>0 a < 0 and b ≥ 0 a < 0 and b < 0 x = 0 and y > 0 x = 0 and y < 0 x = 0 and y = 0 (Quad I and IV) (Quad II) (Quad III) (Upper imag axis) (Lower imag axis) (The origin) Examples Find the modulus r and argument θ for the following numbers (Hint: It is easiest to visualize these in the plane): • z = −3: SOLUTION: r = 3 and θ = π • z = 2i: SOLUTION: r = 2 and θ = π/2 √ • z = −1 + i: SOLUTION: r = 2 and θ = tan−1 (−1) + π = − π4 + π = 3π 4 • z = −3 − 2i (Numerical approx from Calculator OK): √ SOLUTION: r = 14 and θ = tan−1 (2/3) − π ≈ 0.588 − π ≈ −2.55 rad 3 3.1 Operations on Complex Numbers The Conjugate of a Complex Number If z = a + bi is a complex number, then its conjugate, denoted by z̄ is a − bi. For example, z = 3 + 5i ⇒ z̄ = 3 − 5i z = i ⇒ z̄ = −i z = 3 ⇒ z̄ = 3 Graphically, the conjugate of a complex number is it’s mirror image across the horizontal axis. If z has magnitude r and argument θ, then z̄ has the same magnitude with a negative argument. 2 3.2 Addition/Subtraction, Multiplication/Division To add (or subtract) two complex numbers, add (or subtract) the real parts and the imaginary parts separately: (a + bi) ± (c + di) = (a + c) ± (b + d)i To multiply, expand it as if you were multiplying polynomials: (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i and simplify using i2 = −1. Note what happens when you multiply a number by its conjugate: z z̄ = (a + bi)(a − bi) = a2 − abi + abi − b2 i2 = a2 + b2 = |z|2 Division by complex numbers z, w: denominator): z w, is defined by translating it to real number division (rationalize the z w̄ z w̄ z = = w ww̄ |w|2 Example: (1 + 2i)(3 + 5i) −7 11 1 + 2i = = + i 3 − 5i 34 34 34 4 The Polar Form of Complex Numbers 4.1 Euler’s Formula Any point on the unit circle can be written as (cos(θ), sin(θ)), which corresponds to the complex number cos(θ) + i sin(θ). It is possible to show the following directly, but we’ll use it as a definition: Definition (Euler’s Formula): eiθ = cos(θ) + i sin(θ). 4.2 Polar Form of a + bi: The polar form is defined as: z = reiθ 4.2.1 where r = |z| = p a2 + b2 θ = arg(z) Examples Given the complex number in a + bi form, give the polar form, and vice-versa: 1. z = 2i SOLUTION: Since r = 2 and θ = π/2, z = 2eiπ/2 2. z = 2e−iπ/3 We recall that cos(π/3) = 1/2 and sin(π/3) = √ 3/2, so z = 2(cos(−π/3) + i sin(−π/3)) = 2(cos(π/3) − i sin(π/3)) = 1 − 5 √ 3i Exponentials and Logs The logarithm of a complex number is easy to compute if the number is in polar form:

ln(a + bi) = ln reiθ = ln(r) + ln eiθ = ln(r) + iθ The logarithm of zero is left undefined (as in the real case). However, we can now compute the log of a negative number:
ln(−1) = ln 1 · eiπ = iπ or the log of i : 3 ln(i) = ln(1) + π π i= i 2 2 Note that the usual rules of exponentiation and logarithms still apply. To exponentiate a number, we convert it to multiplication (a trick we used in Calculus when dealing with things like xx ): ab = eb ln(a) Example, 2i = ei ln(2) = cos(ln(2)) + i sin(ln(2)) √ i π/4
1/2 √ Example: 1 + i = (1 + i)1/2 = 2e = (21/4 )ei π/8 Example: ii = ei ln(i) = ei(iπ/2) = e−π/2 6 Real Polynomials and Complex Numbers If ax2 + bx + c = 0, then the solutions come from the quadratic formula: √ −b ± b2 − 4ac x= 2a In the past, we only took real roots. Now we can use complex roots. For example, the roots of x2 + 1 = 0 are x = i and x = −i. Check: (x − i)(x + i) = x2 + xi − xi − i2 = x2 + 1 Some facts about polynomials when we allow complex roots: 1. An nth degree polynomial can always be factored into n roots. (Unlike if we only have real roots!) This is the Fundamental Theorem of Algebra. 2. If a + bi is a root to a real polynomial, then a − bi must also be a root. This is sometimes referred to as “roots must come in conjugate pairs”. 7 Exercises 1. Suppose the roots to a cubic polynomial are a = 3, b = 1−2i and c = 1+2i. Compute (x−a)(x−b)(x−c). 2. Find the roots to x2 − 2x + 10. Write them in polar form. 3. Show that: Re(z) = z + z̄ 2 Im(z) = z − z̄ 2i 4. For the following, let z1 = −3 + 2i, z2 = −4i (a) Compute z1 z̄2 , z2 /z1 (b) Write z1 and z2 in polar form. 5. In each problem, rewrite each of the following in the form a + bi: (a) e1+2i (b) e2−3i (c) eiπ (d) 21−i π (e) e2− 2 i (f) π i 6. For fun, compute the logarithm of each number: (a) ln(−3) (b) ln(−1 + i) (c) ln(2e3i ) 4