Overview of Complex Numbers - Study Guide 2022

Overview of Complex Numbers 1 Initial Definitions Definition 1 The complex number z is defined as: z = a + bi, where a, b are real numbers and i = √ −1. General notes about z = a + bi • Engineers typically use j instead of i. • Examples of complex numbers: 5 + 2i, 3− √ 2i, 3, −5i • Powers of i are cyclic: i2 = −1, i3 = −i, i4 = 1, i5 = i, i6 = −1 and so on. Notice that the cycle is: i, −1, −i, 1, then it repeats. • All real numbers are also complex (by taking b = 0), so the set of real numbers is a subset of the complex numbers. We can split up a complex number by using the real part and the imaginary part of the number z: Definition: The real part of z = a + bi is a, or in notation we write: Re(z) = Re(a + bi) = a The imaginary part of a + bi is b, or in notation we write: Im(z) = Im(a + bi) = b 2 Visualizing Complex Numbers 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 the figure to the right, 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) r, and angle (or argument) θ as: p θ = arg(z) r = |z| = a2 + b2 Once we do that, we can write: z = a + bi = r(cos(θ) + i sin(θ)) We have to be a bit careful defining θ so that the question is well-posed. We can define the argument θ as the following, which looks more complicated than it actually is. Highly recommended: Draw the point a + ib in the complex plane.  not defined if (a, b) = (0, 0)     tan−1 (b/a) if (a, b) ∈ QI or QIV  π/2 if a = 0, b > 0 arg(z) = arg(a + ib) = θ =   −π/2 if a = 0, b < 0    tan−1 (b/a) + π if (a, b) ∈ QII or QIII 1 Examples Find the modulus r and argument θ for the following numbers, then express z in polar form: • z = −3: SOLUTION: r = 3 and θ = π so z = 3(cos(π) + i sin(π)) • z = 2i: SOLUTION: r = 2 and θ = π/2 so z = 2(cos(π/2) + i sin(π/2)) • z = −1 + i: SOLUTION: r = √ 2 and θ = tan−1 (−1) + π = − π4 + π = 3π 4 so      √ 3π 3π z = 2 cos + i sin 4 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, or √ √ z = 14 (cos(−2.55) + i sin(−2.55)) = 14(cos(2.55) − i sin(2.55)) Note to readers: We used the “even” symmetry of the cosine and the “odd” symmetry of the sine to do the simplification: cos(−x) = cos(x) 3 3.1 and sin(−x) = − sin(x) 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. EXAMPLE: If z = 3(cos(π/2) + i sin(π/2)), find the conjugate z̄: z̄ = 3(cos(−π/2) + i sin(−π/2)) = 3(cos(π/2) − i sin(π/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. This is like adding polynomials (with i in place of x): (a + bi) ± (c + di) = (a + c) ± (b + d)i To multiply, expand it as if you were multiplying polynomials, with i in place of x: (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i and simplify using i2 = −1. A special product is often computed- A complex number with its conjugate: z z̄ = (a + bi)(a − bi) = a2 − abi + abi − b2 i2 = a2 + b2 = |z|2 2 Division by complex numbers wz , is defined by making it real number division- this is done by rationalizing the denominator using the conjugate of the denominator: z z w̄ z w̄ = = w ww̄ |w|2 Example: −7 11 1 + 2i (1 + 2i)(3 + 5i) (1 + 2i)(3 + 5i) = = = + i 3 − 5i (3 − 5i)(3 + 5i) 32 + 52 34 34 4 The Polar Form of Complex Numbers The polar form of a complex number, z = r cos(θ) + ir sin(θ) has a beautiful counterpart using the complex exponential function, eiθ . First, we’ll define it using Euler’s formula (although it is possible to prove Euler’s formula). Definition (Euler’s Formula): eiθ = cos(θ) + i sin(θ). We can now express the polar form of a complex number slightly differently: p z = reiθ where r = |z| = a2 + b2 θ = arg(z) An important note about this expression: The rules of exponentiation still apply in the complex case. For example,
n ea+ib = ea eib and eiθ eiβ = e(θ+β)i and eiθ = einθ Furthermore, in the next section, we’ll look at the logarithm. 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 SOLUTION: 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. We use the normal rule of logs: ln(ab) = ln(a) + ln(b), or in the case of polar form:

ln(a + bi) = ln reiθ = ln(r) + ln eiθ = ln(r) + iθ Where we leave the last step as intuitively clear, but we don’t prove it here (we have to be careful about the choice of θ as described earlier). 3 The logarithm of zero is left undefined (as in the real case). However, we can now compute things like the log of a negative number!
ln(−1) = ln 1 · eiπ = iπ or the log of i : ln(i) = ln(1) + π π i= i 2 2 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) Examples of Exponentiation √ 2i = ei ln(2) = cos(ln(2)) + i sin(ln(2)), 6 1 + i = (1 + i)1/2 = √ 2ei π/4 1/2 = (21/4 )ei π/8 Real Polynomials and Complex Numbers If ax2 + bx + c = 0, then the solutions come from the quadratic formula: √ −b ± b2 − 4ac x= 2a If the discriminant is negative, then the roots are complex conjugate pairs. NOTE: It is usually easier and quicker to “complete the square” than it is to use the full quadratic formula. 7 Exercises 1. For each polynomial below, first find the roots by completing the square, then write the roots in polar form: (a) x2 − 2x + 10 (b) x2 + 4x + 5. 2. Show that: Re(z) = z + z̄ 2 (c) x2 − x + 1 Im(z) = z − z̄ 2i 3. For the following, let z1 = −3 + 2i, z2 = −4i (a) Compute z1 z̄2 , z2 /z1 (b) Write z1 and z2 in polar form. 4. In each problem, rewrite each of the following in the form a + bi: π (a) e1+2i (c) eiπ (e) e2− 2 i (b) e2−3i (d) 21−i (f) π i 5. For fun, compute the logarithm of each number: (a) ln(−3) (b) ln(−1 + i) 4 (c) ln(2e3i )