Math 147 — Complex Analysis

Home , Complex multiplication, Imaginary unit

Neil Donaldson

Fall 2020

Complex Analysis and Applications, James Ward Brown & Ruel V. Churchill, 9th Ed 2014, McGraw Hill.

1.1 Basic Algebraic Properties (roughly § 1–6 in textbook) There are many ways to introduce the complex numbers, though essentially all center on the seem- ingly absurd equation x2 = 1. Positing a solution to this equation and playing with the result − provided the ﬁrst ‘modern’ discussion by the Italian mathematician Rafael Bombelli1 in the 1500’s.

Definition 1.1. The complex numbers C comprises the vector space R2 together with a new operation (complex multiplication.. Specifically, if z = (x, y) and w = (u, v) are complex numbers (x, y, u, v R), ∈ we define

Addition z + w := (x + u, y + v)

Multiplication zw := (xu yv, xv + yu) − The co-ordinates of a complex number z = (x, y) are termed its real and imaginary parts:

Re z = x, Im z = y

The vector space notation is unwieldy: instead we write

z = x + iy by introducing the symbol i as a short-hand for the point (0, 1): this is termed the imaginary unit. In this notation, the complex number 1 is really the point (1, 0).

It is the strange properties of i under multiplication that make C different (and more interesting!) than R2. Before considering this, we remind ourselves of several simple properties of R2 and how these provide interpretations of the complex numbers.

1While Bombelli was happy to calculate and discuss the relationships between solutions of various quadratic equations, he always considered them entirely ‘ﬁctitious.’

1 Vector Addition Since complex addition is merely the addition of real vectors, we immediately observe that addition of complex numbers is:

Associative: z1 + (z2 + z3) = (z1 + z2) + z3 Commutative: z + w = w + z Indeed commutativity is the familiar parallelogram law from vector spaces: z + w

w = u + iv iv z = x + iy iy

x u These can be proved algebraically, though it is tedious. Instead here’s an easy example of addition in the new language: observe how we simply sum the real and imaginary parts:

Example 1.2. If z = 3 + 4i and w = 2 7i, then − z w = (3 + 4i) (2 7i) = (3 2) + (4 ( 7))i = 1 + 11i − − − − − − When representing the complex numbers, the plane is often known as the Argand diagram. The hor- izontal axis is known as the real axis while the vertical is the imaginary axis. As we’ve done in the picture, it is common to label entries on the imaginary axis using i’s.

The Modulus of a Complex Number & the Triangle Inequality

Deﬁnition 1.3. = + The modulus of a complex number z x iy is the Eu- 2i z = 1 + √3i clidean distance of the point (x, y) from the origin: q z := x2 + y2 i z = 2 | | | | In the picture, z = 1 + √3i has modulus z = √1 + 3 = 2. | | 1 2

Considering the picture at the top of the page, we immediately see the triangle inequality: z + w z + w | | ≤ | | | | Unlike in real analysis, this follows from an honest triangle! By induction, it holds generally z + z + + z z + + z | 1 2 ··· n| ≤ | 1| ··· | n| In the exercises below, you’ll be asked to prove a useful generalization:

z + w z w | | ≥ | | − | |

2 One can use the modulus to efﬁciently describe various curves and regions in the plane:

Examples 1.4. 1. z = 4 describes the circle centered at the origin of radius 4. | | 2. z 3i 2 describes the disk centered at 3i with radius 2. | − | ≤ 3. z + z 1 = 3 describes an ellipse with foci 0 and 1 (the sum of the distances from two ﬁxed | | | − | points is constant). You can try writing these using square-roots and multiplying out to obtain the ‘usual’ algebraic equations: be careful with example 3!

Complex multiplication The link with the equation x2 = 1 comes immediately from the deﬁnition of complex multi- − plication. In what follows, it is important to note that 1, as a complex number is really the point − ( 1, 0) R2! − ∈ Lemma 1.5. i2 = 1. −

Proof.i 2 = (0,1 ) (0,1 ) = (0 0 1 1,0 1 + 1 0) = ( 1, 0) = 1. · · − · · · − −

Now that we’ve established the fundamental property of i, we no longer have any need for the vector space notation: from now on, we will always denote complex numbers in the form z = x + iy.

The basic algebraic properties of multiplication are also straightforward: check one or two by multiplying out!

Lemma 1.6. For any complex numbers z1, z2, z3, we have

Associativity: z1(z2z3) = (z1z2)z3

Commutativity: z1z2 = z2z1

Distributivity: z1(z2 + z3) = z1z2 + z1z3

The upshot of all this is that we can treat complex addition, subtraction and multiplication as if we are working with linear polynomials in the abstract variable i: we simply need to remember to replace i2 with 1 whenever necessary. − Example 1.7. As above, if z = 3 + 4i and w = 2 7i, − zw = (3 + 4i)(2 7i) = 3 2 + 4i 2 3 7i 4i 7i = 6 + 8i 21i 28i2 − · · − · − · − − = 6 + 8i 21i + 28 = 34 13i − −

3 Division of Complex Numbers and the Complex Conjugate

w 1 Division is simply multiplication by an inverse: z = wz− . The inverse of a complex number z = 1 1 x + iy should be some z− = u + iv satisfying zz− = 1. Let us unpack this: ( 1 xu yv = 1 zz− = xu yv + i(xv + yu) = 1 − − ⇐⇒ xv + yu = 0

It is a little tedious, but we can work through this in cases:

• If x = 0, then the second equation require v = y u. Substituting into the ﬁrst equation forces 6 − x y2 x y 1 = xu + u = u = and v = − ( ) x ⇒ x2 + y2 x2 + y2 ∗

• If x = 0, we’re forced to have u = 0 and v = 1 , which ﬁts the same pattern as ( )! − y ∗ In particular, every non-zero complex number z = x + iy has a unique multiplicative inverse:

1 x iy z− = − x2 + y2

This is a critical part of the statement that the complex numbers form a field. Note how the numerator looks very like z itself, except for a negative sign: also observe how the denominator is simply z 2. | | Definition 1.8. The conjugate of a complex number z = x + iy is the complex number z = x iy, − obtained by reflecting z in the real axis.

As observed above, zz = z 2 and, provided z = 0, | | 6

1 z z− = z 2 | | The conjugate helps to facilitate the computation of division:

Example 1.9. If z = 3 + 4i and w = 2 7i, then we computea −

w 2 7i 1 wz (2 7i)(3 4i) 2 3 2 4i 7i 3 + 7i 4i = − = wz− = = − − = · − · − · · z 3 + 4i z 2 3 + 4i 2 32 + 42 | | | | 6 8i 21i + 28i2 22 29i = − − = − − 25 25 aThis can be viewed as multiplying the top and bottom by the conjugate of the denominator:

2 7i 2 7i 3 4i − = − − = 3 + 4i 3 + 4i · 3 4i ··· −

4 Exercises. 1.1.1. Prove that Re(iz) = Im z and that Im(iz) = Re z. − 1.1.2. (a) Check explicitly that z = 2 + 3i and its conjugate z = 2 3i solve the quadratic equation − z2 4z + 13 = 0. − 2 b+i√ω (b) Suppose a, b, c R where ω := 4ac b > 0. Check that z = − and its conjugate z ∈ − 2a both solve the quadratic equation az2 + bz + c = 0. (Since i2 = 1, it makes sense to write √ ω = i√ω: we see that the quadratic formula now − − applies to all real quadratics) 1.1.3. Explicitly prove the commutativity of complex multiplication (Lemma 1.6) using the vector deﬁnition of C (Deﬁnition 1.1). 1.1.4. Evaluate the following in the form x + iy:

2 i 4 2 2 (a) − (b) (1 + i) (c) (2 + 3i)− (2 3i)− 3 5i − − − 1.1.5. Prove the following: you should write z = x + iy rather than using the vector deﬁnition. (a) z = z 1 1 (b) (z− )− = z (c) zw = z w · 1.1.6. (a) For any z, w, use the triangle inequality to prove that z + w z w . | | ≥ | | − | | (b) What relationship between z, w corresponds to equality here? Draw a picture to try to ﬁgure it out! 1.1.7. Suppose that z 2 and consider the polynomial P(z) = z3 + 3z 1. | | ≥ − 3z 1 7 (a) Prove that −3 z ≤ 8 3 3 3z 1 (b) Write P(z) = z + 3z 1 = z 1 + −3 and use the result of Exercise 1.1.6. (a) to | | − z prove that P(z) 1. | | ≥ (This shows that all zeros of P(z) must lie inside the circle z < 2.) | | 1.1.8. By considering the inequality ( x y )2 0, prove that | | − | | ≥ √2 z Re z + Im z | | ≥ | | | | 1.1.9. Prove that the hyperbola x2 y2 = 1 can be written in the form z2 + z2 = 2. − 1.1.10. Draw a picture of the ellipse satisfying the equation z + z 4i = 6. Find the equation of (x c)2 (y d)2 | | | − | − − the curve in Cartesian coordinates: a2 + b2 = 1 where (c, d) is the center of the ellipse and a, b are the semi-axes. (Hint: write z 4i = 6 z , square both sides, cancel x2, y2 terms and repeat. . . ) | − | − | |

5 1.2 The Exponential or Polar Form of a Complex Number (§7–9) Recall Deﬁnition 1.3 of the modulus of a complex number. We extend this to also consider the angle.

Deﬁnition 1.10. A complex number can be written in polar co- 2i = + √ ordinates: z 1 3i

z = x + iy = r cos θ + ir sin θ = r(cos θ + i sin θ) i z = 2 Clearly r = z is the modulus of z. The argument arg z = θ is the angle | | | | π (in radians) measured counter-clockwise from the positive real axis. arg(z) = 3 Its principal value Arg z is chosen such that π < Arg z π. − ≤ √ π 12 In the picture, z = 1 + 3i has argument arg z = 3 .

As with standard polar co-ordinates, the argument can be awkward to calculate: it may seem that the following is all we need ( x = r cos θ y 1 y = tan θ = = arg z = tan− y = r sin θ ⇒ x ⇒ x

However, this only makes sense if x = 0. π 6 2 Moreover, tan 1 has range ( π , π ) and so an ad- − − 2 2 dition or subtraction of π may be required to get the correct value. Here is a picture to remind you, with the principal value Arg z indicated: for instance, z = 3 3i Q2: tan 1 y + π Q1: tan 1 y − − − x − x lies in the third quadrant, so

1 3 3π Arg z = tan− − π = 3 − − 4 π 0 − Since the argument itself is unconstrained, it is perfectly acceptable, for instance, to say that 1 y 1 y 5π Q3: tan− x π Q4: tan− x arg z = 4 . − Note particularly that 0 does not have an argument!

π − 2 Deﬁnition 1.11. If θ R then the exponential eiθ is deﬁned by Euler’s formula ∈ eiθ := cos θ + i sin θ

The polar form of a complex number can now be written z = reiθ where r = z and θ = arg(z). | | If w = u + iv is complex, then its exponential is ew = eueiv = eu(cos v + i sin v).

6 Why should Euler’s formula be true? There are several reasons why Euler’s formula provides a sensible deﬁnition of eiθ.

kθ 1. e is deﬁned as the solution to the initial value problem y0 = ky with y(0) = 1. If we assume differentiation still works when k = i, we see that Euler’s formula satisﬁes this criterion.

2. Evaluate the Maclaurin series for exp z when z = iθ: the real and imaginary parts are easily seen to be the Maclaurin series for cos θ and sin θ.

3. The multiple-angle formulæ for sine and cosine conﬁrm the exponential law

ei(θ+φ) = eiθeiφ (†)

Just as with standard polar co-ordinates, the polar form of a complex number is well-suited to de- scribing circles. For example, the circle centered at z0 = 3 + 4i with radius R = 2 may be parametrized by

iθ iθ z = z0 + Re = 3 + 4i + 2e = 3 + 2 cos θ + i(4 + 2 sin θ)

We thus have a curve parametrized by θ.

Exponential Laws As observed in (†), the complex exponential follows the usual exponential law. In particular we observe that if z = reiθ and w = seiψ, then

zw = rsei(θ+ψ) = zw = z w and arg(zw) = arg z + arg w ⇒ | | | | | | The principal value might not behave so nicely since Arg z = arg z + 2πn for some n Z. ∈ Example 1.12. Find the modulus and argument of zw given z = 7 + i and w = 4 + 3i. We can − − proceed two ways:

1. Find the polar forms of z, w

iθ iψ 1 1 1 3 z = 5√2e and w = 5e where θ = π tan− and ψ = π tan− − 7 − 4 from which

1 1 1 3 zw = z w = 25√2, arg(zw) = arg z + arg w = 2π tan− tan− | | | | | | − 7 − 4

2. First ﬁnd zw = ( 7 + i)( 4 + 3i) = 25 25i then compute its polar form: − − − πi/4 π zw = 25√2e− = zw = 25√2, arg(zw) = ⇒ | | − 4

The discrepancya in the argument comes from the extra copy of 2π: in fact Arg z = π . − 4 a 1 1 1 3 π tan α+tan β tan− 7 + tan− 4 = 4 can be checked using the multiple-angle formula for tangent: tan(α + β) = 1 tan α tan β −

7 The polar form also works easily with division and (integer) exponentiation:

( z r i(θ ψ) = e − z = reiθ, w = seiψ = w s ⇒ zn = rneinθ, n Z ∈

Examples 1.13. 1. The identity (eiθ)n = einθ, equivalently (cos θ + i sin θ)n = cos nθ + i sin nθ is known as de Moivre’s formula; it yields some interesting trig identities. For instance, by taking real parts,

cos 3θ + i sin 3θ = (cos θ + i sin θ)3 = cos3θ + 3i cos2θ sin θ 3 cos θ sin2θ i sin3θ − − = cos 3θ = cos3θ 3 cos θ sin2θ = 4 cos3θ 3 cos θ ⇒ − − 2. We compute z10 when z = √3 i. First observe that z = 2e πi/6, from which − − 10 10 5πi/3 πi/3 z = 2 e− = 1024e = 512(1 + √3i)

Exercises. 1.2.1. Use induction to prove that for any n N 2 we have ∈ ≥ iθ iθ iθ i(θ +θ + +θ ) e 1 e 2 e n = e 1 2 ··· n ··· 1.2.2. Find the principal argument of (1 + i)2020. iθ iθ iθ 1.2.3. Prove that e = 1 and that e = e− . 1.2.4. Show that if Re z > 0 and Re w > 0, then Arg(zw) = Arg z + Arg w. 1.2.5. Use de Moivre’s formula to establish the identity

cos 4θ = 8 cos4θ 8 cos2θ + 1 − 1 3 1 1 1.2.6. With reference to Example 1.12, if α = tan− 4 and β = tan− 7 , show that 4 3 7 1 cos α = , sin α = , cos β = , sin β = 5 5 √50 √50

π Now use the cosine multiple-angle formula to check that α + β = 4 . For a challenge, generalize your approach to prove the general multiple-angle formula for tangent, at least when α, β are acute angles.

8 1.3 Roots of Complex Numbers (§10–11) Taking roots is a difﬁcult proposition in C. For instance, consider squaring c = 2 + 3i, which has

z := c2 = 5 + 12i − If we were given z = 5 + 12i, how would we go about ﬁnding a square-root c? Naively, you might − try hacking at it: ( x2 y2 = 5 5 + 12i = (x + iy)2 = x2 y2 + 2ixy − − − − ⇐⇒ xy = 6

1 2 Substituting y = 6x− into the ﬁrst equation yields a quadratic in x :

x4 + 5x2 36 = (x2 4)(x2 + 9) − − from which we’d conclude x = 2 and obtain the square-roots c = (2 + 3i). This is messy and ± ± ± hard to generalize; imagine if we were working with cube, or higher, roots!

This is where the polar form comes in. Suppose n N and that c, z satisfy z = cn. In polar form ∈ z = reiθ, c = seiψ = reiθ = sneinψ ⇒ By equating moduli and arguments, we conclude that

r = sn nψ = θ + 2πk ( ) ∗ where k is any integer. We can put this together to obtain a proper deﬁnition; since there are some conventions to follow however, we ﬁrst do an example:

Example 1.14. = 1.5 πi/3 has ﬁfth roots ,..., : z z e c0 c4 c1 5 πi/15 c0 = √1.5e 5 πi/15+2π/5 5 7πi/15 r c1 = √1.5e = √1.5e 5 c0 5 πi/15+4π/5 5 13πi/15 c √r c2 = √1.5e = √1.5e 2 θ θ 5 √5 πi/15+6π/5 √5 19πi/15 c3 = 1.5e = 1.5e 2π 5 πi/15+8π/5 5 25πi/15 5 c4 = √1.5e = √1.5e

[ ) We’ve chosen arguments in the range 0, 2π rather than principal c arguments. Note how the roots form the vertices of a regular pen- 4 5 c tagon, equally spaced around the circle of radius √1.5. Note also 3 how there are precisely ﬁve 5th roots: once k n in ( ), the roots ≥ ∗ start repeating.

9 Deﬁnition 1.15. Given z = reiθ and a positive integer n, the nth roots of z are the n complex numbers

(θ + 2kπ)i c = √n r exp k = 0, 1, . . . , n 1 k n − The nth roots form the vertices of a regular n-gon, equally spaced around the circle of radius √n r. There are some conventions on how to refer to nth roots: suppose θ = Arg z is the principal argument. We write:

√n z = √n reiθ/n: this is the principal nth root of z.

1/n th z = c0,..., cn 1 : this is the set of n roots of z. { − } The nth roots of unity are commonly encountered: these are the set of nth roots of 1 C: ∈ 11/n = e2πki/n : k = 0, . . . , n 1 { − } The notation ω = e2πi/n is often seen, so that the roots of unity are the set ωk : k = 0, . . . , n 1 . n { − } Compare this with what happens in the real numbers. If r = 16, we might write

√4 16 = 2 for the principal fourth root.

161/4 = 2 for all (real) fourth roots. In complex analysis, there are four fourth roots, and we’d ± write 161/4 = 2, 2i, 2, 2i . { − − } Depending on the format (rectangular versus polar), it can be very hard to explicitly compute nth roots 1 of a complex number, particularly if arg z is only known in tan− format. Various trig identities can be used, but in practice this might require more effort than it is worth.

Example 1.16. Find the fourth roots of z = 1 + i. √ πi/4 π First we write in polar form: z = 2e . Since Arg z = 4 , we see that the principal fourth root is √4 √8 πi/16 π 1 + i = 2e . To evaluate this in rectangular form requires us to compute cosine and sine of 16 . For this, the half-angle formulas are useful: s s α 1 π 1 1 √2 + 1 1q cos2 = (1 + cos α) = cos = 1 + = = 1 + √2 2 2 ⇒ 8 2 √2 4 2

Continuing this, and applying similar expressions for sine, we obtain: r r π π √2 8 q √2 8 q √4 1 + i = √8 2 cos + i√8 2 sin = 2 + 1 + √2 + i 2 1 + √2 16 16 2 2 − This is deﬁnitely not attractive! The set of fourth roots is then the product of this with each of the fourth roots of unity: 11/4 = 1, ω , ω2, ω3 = 1, i, 1, i . In this example, it is probably not worth { 4 4 4} { − − } the effort of converting to rectangular form.

10 Exercises. 1.3.1. Find the square roots of √3 + i and express them in rectangular co-ordinates. − (Hint: you may find it useful that (√3 1)2 = 4 2√3... ) − − 1.3.2. Find the sixth roots of i in polar co-ordinates. Which is the principal root? 1+√3i 2 1 √3i 1.3.3. Use the fact that the cube roots of unity are 1, ω3 = − 2 and ω3 = − −2 to evaluate the cube roots of 27 in rectangular co-ordinates. − 1.3.4. We previously found the fourth roots of 16. Use these to find the fourth roots of 16. Hence − factorize the equation z4 + 16 = 0 as a product of two quadratic equations with real coeffi- cients. n 1 1.3.5. If ω is any nth root of unity other than 1, prove that ∑− ωk = 0. k=0 (Hint: recall geometric series) 1.3.6. (a) Suppose that a, b, c C with a = 0 and suppose that z satisfies the quadratic equation ∈ 6 az2 + bz + c = 0. Prove the quadratic formula:

b + (b2 4ac)1/2 z = − − 2a

Note that (b2 4ac)1/2 is the set of square roots of b2 4ac, so that this provides two − − solutions whenever b2 4ac = 0. − 6 (b) Find the roots of the equation iz2 + (1 + i)z + 3 = 0 in rectangular form.