COMPLEX NUMBERS LAURENT DEMONET

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1 COMPLEX NUMBERS LAURENT DEMONET Contents Introduction: numbers 1 1. Construction of complex numbers. Conjugation 4 3. Geometric representation of complex numbers The unit circle Modulus of a complex number Argument of a complex number 9 4. Solving equations in C n-th roots Quadratic equations The fundamental theorem of algebra 18 Note: the proofs with a have to be understood. The other proofs are here for convenience. Introduction: numbers Let us summarize the sets of numbers we already know: We know natural numbers: N t0, 1,,... u. It turns out that the equation x ` 1 0 has no solution. So: We introduce integers: Z t...,, 1, 0, 1,,... u. Here, x`1 0 has a solution: 1. But 3x 1 has no! solution. So: a We introduce rational numbers: Q b a P Z, b P N Nzt0u). Here, 3x 1 has a solution: 1{3. But x has no solution. Indeed, suppose that there is a rational number a{b such that pa{bq. Up to simplification, we can suppose that a or b is odd (or maybe both). We deduce from pa{bq that a b. From that, we get that a is an even number. As a is even, then a is also even. And therefore, we can write a ˆ a 1 for another integer a 1. Continuing the computation, we get pa 1 q b so 4a 1 b so a 1 b and therefore b is even. Finally, b is also even and it contradicts the fact that one at least of a and b has to be odd. So: We introduce real numbers: R tpossible coordinates of a lineu. This definition is not the most rigorous possible. The rigorous definition is outside of the scope of this course. In R, the equation 1

2 LAURENT DEMONET x has two solutions,? and?. But x 1 has no solutions. We will now construct a bigger set of numbers, called complex numbers where x 1 has a solution. Indeed, we will see at the end that in this set of numbers, every polynomial equation (that is an equation of the form a n x n `a n 1 x n 1 ` `a 1 x`a 0 ) has at least a solution (and, in some sense, n solutions where n is the degree of the polynomial). 1. Construction of complex numbers Definition 1.1. We denote C R. We add a multiplication ˆ on C by defining, for all pa, bq, pa 1, b 1 q P C, pa, bq ˆ pa 1, b 1 q paa 1 bb 1, ab 1 ` ba 1 q. Thus, we get a set of number C. Indeed, we already had an addition and a subtraction because C R is a vector space (over R). Notation 1.. Often, we will use the following two symbols: D The first one has to be read as there exists and the second one as for all. Thus, we can rewrite the definition of the multiplication in C bq, pa 1, b 1 q P C, pa, bq ˆ pa 1, b 1 q paa 1 bb 1, ab 1 ` ba 1 q. Example 1.3. The elements p1, q and p3, 4q are two elements of C. We have p1, q ˆ p3, 4q p1 ˆ 3 ˆ 4, 1 ˆ 4 ` ˆ 3q p 5, 10q. Proposition 1.4. For any x, y, z P C, we have px ˆ yq ˆ z x ˆ py ˆ zq. Moreover, x ˆ y y ˆ x. Proof (*). Let x, y, z P C. As complex numbers, x, y and z can be rewritten as x pa, bq, y pa 1, b 1 q and z pa, b q where a, b, a 1, b 1, a and b are real numbers. Thus, we have px ˆ yq ˆ z ppa, bq ˆ pa 1, b 1 qq ˆ pa, b q paa 1 bb 1, ab 1 ` ba 1 q ˆ pa, b q and ppaa 1 bb 1 qa pab 1 ` ba 1 qb, paa 1 bb 1 qb ` pab 1 ` ba 1 qa q paa 1 a bb 1 a ab 1 b a 1 bb, aa 1 b bb 1 b ` ab 1 a ` ba 1 a q x ˆ py ˆ zq pa, bq ˆ ppa 1, b 1 q ˆ pa, b qq pa, bq ˆ pa 1 a b 1 b, a 1 b ` b 1 a q papa 1 a b 1 b q bpa 1 b ` b 1 a q, apa 1 b ` b 1 a q ` bpa 1 a b 1 b qq paa 1 a ab 1 b ba 1 b bb 1 a, aa 1 b ` ab 1 a ` ba 1 a bb 1 b q so px ˆ yq ˆ z x ˆ py ˆ zq. For the second part of the proposition, compute: x ˆ y pa, bq ˆ pa 1, b 1 q paa 1 bb 1, ab 1 ` ba 1 q and which are clearly equal. y ˆ x pa 1, b 1 q ˆ pa, bq pa 1 a b 1 b, a 1 b ` b 1 aq Remark 1.5. Because of Proposition 1.4, we write xˆyˆz instead of pxˆyqˆz or x ˆ py ˆ zq.

3 COMPLEX NUMBERS 3 Proposition 1.6. There is a unique f P C such that, for any x P C, f ˆx x ˆ f x. We denote 1 f. If x P C is different from p0, 0q, there exists a unique x 1 such that xˆx 1 x 1 ˆ x 1. We denote x 1 x 1. Proof. We want to prove that there is a unique f P C such that, for any x P C, f ˆ x x ˆ f x. To do this kind of proof, we first have to find f P C which satisfies this requirement. It turns out that f p1, 0q satisfies it. Indeed, for any x P C, we can write x pa, bq with a, b P R and f ˆ x p1, 0q ˆ pa, bq p1 ˆ a 0 ˆ b, 1 ˆ b ` 0 ˆ aq pa, bq and x ˆ f f ˆ x x thanks to Proposition 1.4. The second step of the proof consists to show that f is unique, which means that there is no other choice for f. Suppose that f 1 is another complex number such that for any x P C, f 1 ˆ x x ˆ f 1 x. Let us study f ˆ f 1. By the property of f, f ˆ f 1 f 1. By the property of f 1, f ˆ f 1 f. From that, we deduce that f f 1 and therefore, there is no other choice than f. The second part of the proposition states that for any x P Czp0, 0q there exists a unique x 1 such that x ˆ x 1 x 1 ˆ x 1. So suppose that x P C. We can write x pa, bq where a, b P R. Once again, the first step consists to find x 1 which satisfies the requirement. It turns out that the complex number x 1 pa{pa ` b q, b{pa ` b qq satisfies the requirement. Indeed ˆ a x ˆ x 1 pa, bq ˆ b ˆ a ˆ a ` b, a ` b a a ` b b ˆ b a ` b, a ˆ b a ` b ` b ˆ p1, 0q 1 ˆa ` b a ` b, a ˆ b ` b ˆ a a ` b a a ` b and x 1 ˆ x x ˆ x 1 1 by Proposition 1.4. Moreover, suppose that we have another x P C such that x ˆ x x ˆ x 1. Let us compute x 1 ˆ x ˆ x in two different ways. First, we have, thanks to Proposition 1.4, Secondly, we have, x 1 ˆ x ˆ x px 1 ˆ xq ˆ x 1 ˆ x x. x 1 ˆ x ˆ x x 1 ˆ px ˆ x q x 1 ˆ 1 x 1 so x 1 x which proves that there is no other x. Example 1.7. Let z p1, 3q. Then ˆ 1 z 1 1 ` 3, 3 1 ` 3 and we can check that ˆ 1 z ˆ z 1 p1, 3q ˆ 10, 3 10 p1, 0q 1. ˆ 1 10, 3 10 ˆ 1 ˆ 1 10 ` 3 ˆ 3 10, 1 ˆ 3 10 ` 3 ˆ 1 10

4 4 LAURENT DEMONET Notation 1.8. If x, y P C and y p0, 0q, we denote x{y x ˆ y 1. Remark that x y ˆ y y ˆ x y x. Remark 1.9. The formula to compute x{y and y 1 for x P C and y P Czt0u has to be known. From now on, we denote i p0, 1q P C and, more generally, a ` bi pa, bq P C. This notation makes sense as we denoted 1 p1, 0q. Notice that pa ` biq ˆ pa 1 ` b 1 iq aa 1 ` ab 1 i ` a 1 bi bb 1 so that the computations in C can be done as usual, adding the rule that i 1. Moreover, it will also happen that we omit the symbol ˆ. Example Let us compute an example: p3 ` iqp5 iq 3 ˆ 5 3 ˆ i ` i ˆ 5 ` i ˆ p iq 15 3i ` 10i i 15 ` 1i ` 17 ` 7i. Definition For a ` bi P C, we call a the real part of a ` bi and b its imaginary part. We denote Repa ` biq a and Impa ` biq b. Example 1.1. Rep5 ` 4iq 5 and Imp5 ` 4iq 4. Pay attention that Imp5 ` 4iq 4i! Proposition The map Re : C Ñ R is linear (over R). Proof (*). To prove that Re is linear, we have to check that for every z, z 1 P C, Repz ` z 1 q Repzq ` Repz 1 q. So, let z, z 1 P C. We know that we can write z a ` bi and z 1 a 1 ` b 1 i for some a, b, a 1, b 1 P R. And then Repz ` z 1 q Repa ` bi ` a 1 ` b 1 iq Reppa ` a 1 q ` pb ` b 1 qiq a ` a 1 Repa ` biq ` Repa 1 ` b 1 iq Repzq ` Repz 1 q. We also have to check that for any λ P R and z P C, Repλzq λ Repzq. So, let λ P R and z P C. We know that we can write z a ` bi for some a, b P R. Then Repλzq Repλpa ` biqq Repλa ` λbiq λa λ Repa ` biq. Finally, Re : C Ñ R is a linear map. Exercise Prove that Im : C Ñ R is linear.. Conjugation Definition.1. Let a ` bi P C. We call the number a ` bi a bi. the conjugate of a ` bi. In other terms, if z P C, its conjugate z satisfies Repzq Repzq and Impzq Impzq. Example.. 5 ` 3i 5 3i. Proposition.3. For any z, z 1 P C, (i) pzq z; (ii) z ` z 1 z ` z 1 ;

5 COMPLEX NUMBERS 5 (iii) z z; (iv) z z 1 z z 1 ; (v) zz 1 z z 1 ; (vi) z 1 z 1 ; z (vii) z 1 z z 1. Proof (*). Let z, z 1 P C. We know that we can write z a`bi and z 1 a 1`b1 i with a, b, a 1, b 1 P R. Thus: (i) pzq `a ` bi a bi a ` bi z. (ii) z ` z 1 pa ` a 1 q ` pb ` b 1 qi pa ` a 1 q pb ` b 1 qi pa biq ` pa 1 b 1 iq z ` z 1. (iii) z pa ` biq a bi a ` bi pa biq a ` bi z. (iv) z z 1 z ` p z 1 q z ` z 1 z z 1. (v) zz 1 pa ` biqpa 1 ` b 1 iq paa 1 bb 1 q ` pab 1 ` ba 1 qi paa 1 bb 1 q pab 1 ` ba 1 qi. Moreover, z z 1 pa biqpa 1 b 1 iq paa 1 p bqp b 1 qq ` pap b 1 q ` p bqa 1 qi paa 1 bb 1 q pab 1 ` ba 1 qi. ˆ a (vi) z 1 a ` b, b a ` b i a a ` b ` b a ` b i a a ` p bq b a ` p bq i pa biq 1 z 1. z (vii) z 1 z ˆ z 1 1 z ˆ z 1 1 z ˆ z 1 1 z z Geometric representation of complex numbers As seen in the previous section, a complex number z a ` bi P C is also a vector pa, bq in R. In particular, the addition of two complex numbers in C corresponds to the usual addition of vectors. Example 3.1. For example, we can add z `.6i and z 3.4 ` i: z 1 ` z 1.8 ` 3.6i z `.6i i z 3.4 ` i 1 Remark 3.. The conjugate of a complex number z corresponds geometrically to its symmetric around the X axis. Indeed, if z pa, bq with a, b P R, then z pa, bq.

6 6 LAURENT DEMONET 3.1. The unit circle. Notation 3.3. If θ P R, we denote e θi cos θ ` psin θqi. Example 3.4. e π 3 i 1? 3 ` i: i e π 3 i π 3 1 Proposition 3.5. The image of the map is the unit circle of R. R Ñ C θ ÞÑ e θi Proof (*). The equation of the unit circle is x ` y 1. On the other hand, if θ P R, e θi cos θ ` sin θi and cos θ ` sin θ 1. So e θi is on the unit circle. i θ 1 e θi px, yq Reciprocally, if px, yq is on the unit circle, x ` yi e θi where θ { p1, 0q, px, yq is the measure of the oriented angle from the vector p1, 0q to the vector px, yq. Proposition 3.6. For all θ, θ 1 P R, e pθ`θ1 qi e θi e θ1i.

7 COMPLEX NUMBERS 7 Proof (*). Let θ and θ 1 be in R. Then e pθ`θ1 qi cospθ ` θ 1 q ` sinpθ ` θ 1 qi pcos θ cos θ 1 sin θ sin θ 1 q ` pcos θ sin θ 1 ` sin θ cos θ 1 qi pcos θ ` psin θqiqpcos θ 1 ` psin θ 1 qiq e θi e θ1 i which finishes the proof. Exercise 3.7. Find all θ P R such that e θi P R. Proposition 3.8. For all θ P R, e θi e θi Proof (*). Let θ P R. By definition, e θi 1. e θi cosp θq ` sinp θqi cos θ ` p sin θqi cos θ psin θqi e θi. Moreover, e θi e θi e pθ`p θqqi e 0i cos 0 ` psin 0qi 1 so, by definition, e θi 1 e θi. Proposition 3.9. For all θ P R and n P Z, e nθi e θi n. Proof (*). Let θ P R and n P Z. If n 0, it is clear that e 0θi e θi 0 1. If n ą 0, then, using Proposition 3.6, e nθi e pθ`pn 1qθqi e θi e pn 1qθi e θi e θi e pn qθi. e θi n. Then, if n ă 0 then n ą 0 and, using Proposition 3.8, e nθi e p nθiq e nθi 1 ˆ e θi n 1 e θi n which finishes the proof. 3.. Modulus of a complex number. Definition If z P C, the modulus of z is z a Repzq ` Impzq. In other terms, it is the distance between 0 and z in the complex plane. Example ` 4i? 3 ` 4? 9 ` 16? 5 5.

8 8 LAURENT DEMONET 3 ` 4i 3 ` 4i 5 i 1 Notation 3.1. The symbol ñ means implies. In other terms, if P and Q are two (mathematical) sentences, P ñ Q means that when P is true then Q is also true. For example, it is true for all x P R that x 3 ñ x but not that x ñ x 3. The symbol ô means if and only if. So P ô Q means that P ñ Q and Q ñ P. For example, it is true P R, x 4 ô x. Proposition Let z, z 1 P C. We have (i) z 0 if and only if z 0; (ii) z z if and only if z P R`; (iii) zz 1 z z 1 ; (iv) if z P R, z is its absolute value; (v) z z ; (vi) z z ; (vii) if z 0, ˇˇz 1ˇˇ z 1 ; (viii) if z 1 0, ˇ z z ˇˇˇ z 1 z 1 ; (ix) z zz; (x) z 1 if and only if z is on the unit circle. Proof (*). Let z, z 1 P C. Let a Repzq, b Impzq, a 1 Repz 1 q and b 1 Impz 1 q. We get: (i) z 0 ô? a ` b 0 ô a ` b 0 ô a b 0 ô z 0. (ii) z z ô? a ` b a ` bi. As? a ` b P R, if it is true then b 0 and? a a so a 0. Finally, in this case z a P R`. Reciprocally, if z P R`, then b 0 and z? a ` b? a ` 0 a z. (iii) We have: zz 1 pa ` biqpa 1 ` b 1 iq paa 1 bb 1 q ` pab 1 ` ba 1 qi a paa 1 bb 1 q ` pab 1 ` ba 1 q a a a 1 aa 1 bb 1 ` b b 1 ` a b 1 ` ab 1 ba 1 ` b a 1 a a a 1 ` b b 1 ` a b 1 ` b a 1 a pa ` b qpa 1 ` b 1 q a a ` b a a 1 ` b 1 z z 1.

9 COMPLEX NUMBERS 9 (iv) If z P R then b 0 and z? a ` b? a which is a if a 0 and a if a 0. It is the absolute value of a z. (v) z a a ` p bq? a ` b z. (vi) z p 1q ˆ z 1 z z. (vii) If z 0, ˇ ˇz 1ˇˇ a ˇa ` b b a ` b i ˇ d a pa ` b q ` d b pa ` b q a ` b pa ` b q 1? a ` b z 1. (viii) If z 1 0, ˇ z ˇˇˇ ˇˇzz 1 1ˇˇ z 1 z ˇˇz 1 1ˇˇ z z 1 1 z z 1. (ix) zz pa ` biqpa biq a pbiq a ` b z. (x) z 1 ô? a ` b 1 ô a ` b 1 ô z pa, bq is on the unit circle Argument of a complex number. Proposition For any z P C, there exists θ P R such that z z e θi. Proof (*). Let z P C. If z 0 then z 0 e 0i for example. We suppose now that z 0. Thanks to Proposition 3.13, we have: z ˇ z ˇ z z z z 1 so, using again Proposition 3.13, z{ z is on the unit circle, and then, by Proposition 3.5, there exists θ P R such that z z eθi so z z e θi. Definition If z P C is not 0, we define Argpzq θ where θ is the real number appearing in Proposition We call Argpzq the argument of z. Moreover, the the form z e Argpzqi is called the polar form of z. Example Suppose that z 1 ` i. Then z? 1 ` 1?. Thus, so Argpzq π{4. z{ z 1? ` i? cos π ` sin 4 π e π 4 i 4

10 10 LAURENT DEMONET z z? 1 `? i i z 1 ` i z? π 4 1 Proposition Let θ P R. Multiplying by e θi in C consists to rotate by θ around 0. Proof (*). Let θ P R and z P C such that z 0. Then e θi z e θi z e Argpzqi z e pθ`argpzqqi. i z Argpzq ` θ θ Argpzq 1 e θi z Of course, it works also for z 0 as e θi 0 0. Proposition Let θ, θ 1 P R. We have e θi e θ1i ô Dn P Z, θ 1 θ nπ. Proof (*). Let θ, θ 1 P R. We have e θ1i e θi ô e θ1i e θi e θi e θi ô e pθ1 θqi 1 ô cospθ 1 θq ` sinpθ 1 θqi 1 ô cospθ 1 θq 1 and sinpθ 1 θq 0 ô Dn P Z, θ 1 θ nπ. It finishes the proof.

11 COMPLEX NUMBERS 11 i θ θ ` π 1 θ π Remark If z is a nonzero complex number then Argpzq is well determined only up to π because of previous Proposition. In other terms, the notation Argpzq is not really appropriate even if it is commonly used. For example, it is valid to write that Argpiq π{ but also Argpiq 5π{ or Argpiq 3π{ as i e π i e 5π i e 3π i. Thus, when we write an equality between arguments (or more generally angles), it has usually to be understood as an equality up to a multiple of π. We will see later in the course that there are ways to partially solve this kind of questions. Proposition 3.0. If z, z 1 P C are nonzero, (i) Argpzq 0 if and only if z P R`; (ii) Argpzz 1 q Argpzq ` Argpz 1 q (up to a multiple of π); (iii) Arg `z 1 Argpzq; (iv) Arg pzq Argpzq; (v) Argp zq Argpzq ` π. Proof (*). Let z and z 1 be two nonzero complex numbers. (i) By definition, Argpzq 0 ô z z e 0i z which is equivalent to say that z P R` by Proposition (ii) By definition, we have zz 1 z e Argpzqi ˆ z 1 e Argpz1 qi zz 1 e pargpzq`argpz1 qqi which implies that Argpzq ` Argpz 1 q corresponds to the definition of Argpzz 1 q. Finally, Argpzz 1 q Argpzq ` Argpz 1 q. (iii) We have Argpzq ` Arg `z 1 Arg `zz 1 Argp1q 0 so Arg `z 1 Argpzq. (iv) We have, thanks to Proposition 3.13, Argpzq ` Arg pzq Arg pzzq Arg ` z 0 because z P R`. So Arg pzq Argpzq.

12 1 LAURENT DEMONET (v) We have Argp zq Argpp 1q ˆ zq Argp 1q ` Argpzq π ` Argpzq. Proposition 3.1 (Euler formulas). Let θ P R. We have Proof (*). We have and cos θ eθi ` e θi e θi ` e θi e θi e θi i which finishes the proof. and sin θ eθi e θi i cos θ ` psin θqi ` cosp θq ` sinp θqi cos θ ` psin θqi ` cos θ psin θqi cos θ cos θ ` psin θqi cosp θq sinp θqi i cos θ ` psin θqi cos θ ` psin θqi i sin θ Proposition 3.. If z P C is nonzero, then z ` z z cos Argpzq and z z z psin Argpzqqi. Exercise 3.3. Using the same technics than in Proposition 3.1, prove Proposition 3. (use the polar form of z). Proposition 3.4 (triangular inequality). If z, z 1 P C, we have ˇ ˇ z z 1 ˇˇ z ` z 1 z ` z 1. Moreover, the first inequality is an equality if and only if Argpzq Argpz 1 q`π or z 0 or z 1 0 and the second one is an equality if and only if Argpzq Argpz 1 q or z 0 or z 1 0. Proof. Let z and z 1 be in C. If z 0 or z 1 0, the result is clear. So let us suppose that z 0 and z 1 0. Let us start by the second inequality. We have, thanks to Propositions 3.13 and 3., z ` z 1 pz ` z 1 qpz ` z 1 q pz ` z 1 qpz ` z 1 q zz ` zz 1 ` z 1 z ` z 1 z 1 z ` pzz 1 q ` zz 1 ` z 1 z ` zz 1 cos Argpzz 1 q ` z 1 z ` z z 1 cos `Argpzq ` Argpz 1 q ` z 1 z ` z z 1 cos `Argpzq Argpz 1 q ` z 1 ` z ` z 1 z z 1 `1 cos `Argpzq Argpz 1 q. We know that 1 cos pargpzq Argpz 1 qq 0 so z ` z 1 p z ` z 1 q and, as z `z 1 0 and z ` z 1 0, we deduce that z `z 1 z ` z 1. Moreover,

13 COMPLEX NUMBERS 13 the equality holds if and only if 1 cos pargpzq Argpz 1 qq 0 which is true if and only if Argpzq Argpz 1 q 0 (up to a multiple of π). For the first equality, we separate two cases: If z z 1. In this case: z ` z 1 p z ` z 1 ` z 1 q z 1 p z ` z 1 ` z 1 q z 1 pz ` z 1 q ` p z 1 q z 1 z z 1 where the inequality of the second line is the one we already proved. In particular, if equality holds, Argpz ` z 1 q Argp z 1 q. Then 0 Argpz ` z 1 q Argp z 1 q Argpz ` z 1 q ` Arg ` z 1 1 ˆ z Arg ` z1 z 1 Arg z 1 z 1 so z{z 1 1 P R` and then z{z 1 is a negative real number. So z π Arg z 1 Argpzq Argpz 1 q. If z 1 z. It is enough to exchange z and z 1 as z z 1 z 1 z. The only thing we did not prove is that if Argpzq Argpz 1 q ` π then z z 1 z ` z 1. For that, just remark that, in this case, z ` z 1 ˇ ˇ z e Argpzqi ` z 1 e pargpzq πqiˇˇˇ Argpzqi ˇˇˇ z e ` z 1 e Argpzqi e πiˇˇˇ ˇ ˇ z e Argpzqi z 1 e Argpzqiˇˇˇ Argpzqi ˇˇˇe ˇ ˇˇ z z 1 ˇˇ ˇˇ z z 1 ˇˇ. It finishes the proof. Remark 3.5. The previous inequalities are called triangular because the second one can be understood as the usual triangular inequality (if we have a triangle then the length of any side is smaller than the sum of the two other lengths): z z ` z 1 i z 1 z 1 z ` z 1 z z 1 z Example 3.6. We can now construct geometrically numbers: 1

14 14 LAURENT DEMONET i e π i z e π i z ` z π 0 1 e π i z In the previous diagram, we constructed from z. e π i z ` 4.1. n-th roots. 4. Solving equations in C Definition 4.1. Let z P C and n P N. We say that z 1 is an n-th root of z if z 1n z. Remark 4.. Recall that if x P R`, we denote by n? x its non negative n-th root. We will never use this notation for numbers which are not in R` in this course. Proposition 4.3. Let z be a nonzero complex number and n P N. Then z has n distinct n-th roots for k P t0, 1,..., n, n 1u. z k na z e Argpzq`πk i n Proof. Let z P Czt0u and n P N. By definition, z z e Argpzqi. Moreover, for any integer k, zk n a Argpzq`πk n z e i n n z e pargpzq`πkqi z e Argpzqi e πki z. Note also that, if k, k 1 P Z, z k z k 1 ô na z e Argpzq`πk n ô e Argpzq`πk n i na z e Argpzq`πk1 i n i e Argpzq`πk1 n i ô e Argpzq`πk i n e Argpzq`πk1 n i 1 ô e pargpzq`πkq pargpzq`πk1 q n i 1 ô e πpk k1 q n i 1 ô Dm P Z, πpk k1 q πm (Prop. 3.18) n ô Dm P Z, k k 1 mn

15 COMPLEX NUMBERS 15 and we deduce from that that all the distinct possible values of z k are taken when k P t0, 1,..., n, n 1u. Suppose now that z 1 is an n-th root of z. Then we have z z 1n z 1 qi n ˇˇz1ˇˇn e Argpz1 e n Argpz1 qi so z z 1 n and therefore, as z 1 is non negative, z 1 na z and, on the other hand, Argpzq n Argpz 1 q up to a multiple of π. So there exists k P Z such that Argpzq ` πk n Argpz 1 q and finally Argpz 1 Argpzq ` πk q n and therefore, z 1 is one of the z k s. There are no other n-th roots. Remark has exactly one n-th root for any n P N : 0 itself. Once again, we never use the notation? i or? 1 for example (at least in this course and its exams). The previous proposition has to be understood in the following way. If we want to compute n-th roots of z z e Argpzqi then we have to take separately the non negative n-root of z and the n-th roots of e Argpzqi. Let us concentrate on this factor. We know that, for any θ P R, e θi n e nθi so, if we want it to be equal to e Argpzqi we have to divide Argpzq by n. The problem that we face then is that, as observed previously, Argpzq is only defined up to multiples of π and therefore Argpzq{n is only defined up to multiples of π{n. But, in general, a multiple of π{n can not be neglected as it is not equivalent to a 0 angle. Argpzq Argpzq ` π Argpzq ` 4π i z 1 In this picture, we have drawn a point and different values of its argument. For each of them we draw the third of the argument. Each time we add π to Argpzq, we add one third of a complete turn around the circle. Thus, it is not necessary to do it a fourth time as we would return to the first possible argument.

16 16 LAURENT DEMONET Note that we call square roots the -nd roots and cube roots the 3-rd roots. Note also that in the case of square roots, the two roots are opposite from each other. Example 4.5. To solve the equation z 3 i, we proceed in the following way. We want to compute cube roots of i so we have to write i in polar form: i e π i. Thus, using the formula of Proposition 4.3, i has 3 cube roots: z 0 3? e π 6 i and z 1 3? e 5π 6 i and z 3? e 9π 6 i which are the three solutions of the equation. Try to draw them in the plane. 4.. Quadratic equations. Definition 4.6. A quadratic equation in C is an equation of the form c z ` c 1 z ` c 0 0 where c, c 1 and c 0 are complex numbers and c 0. Solving this equation means to find all the values of z which satisfy the equation. Proposition 4.7. Let c z ` c 1 z ` c 0 0. equation c 1 4c c 0. We call discriminant of the Denote δ 1 and δ the two square roots of (if 0, then δ 1 δ 0). Then, in general, the equation has two solutions: z 1 c 1 ` δ 1 and z 1 c 1 ` δ. Note that if 0, z 1 z is the unique solution. In this case, we say that this solution is double (as it appears twice). Proof (*). For z P C, we have c z ` c 1 z ` c 0 0 ô z ` c1 c z ` c0 ô ô ˆ z ` c1 ˆ z ` c1 c 0 because c 0 c 1 4c ` c0 0 c 4c If δ 1 and δ are the square roots of, then are the square roots of δ 1 and δ 4c

17 so COMPLEX NUMBERS 17 c z ` c 1 z ` c 0 0 ô z ` c1 δ 1 or z ` c1 δ ô z c 1 ` δ 1 or z c 1 ` δ which is what we wanted to prove. Example 4.8. Let us solve the equation Compute the discriminant: z ` p1 ` iqz ` i 0. p1 ` iq 4 ˆ 1 ˆ i 1 ` i 1 4i i which is nonzero. We have then to compute its two square roots. For that, we should write in polar form. We have a 0 ` p q. Thus i 3π i e i and we deduce that Argp q 3π{. Thus, the two square roots of are δ 1? e 3π 4 i? ˆ cos 3π ˆ 4 ` sin 3π i? ˆ?? 4 ` i 1 ` i. and δ? e 7π 4 i δ 1 1 i. Then the solutions of the quadratic equation are z 1 1 i ` p 1 ` iq 1 and z 1 i ` p1 iq i. Proposition 4.9. If c z ` c 1 z ` c 0 0 is a quadratic equation and z 1, z are its two solutions (with z 1 z if the discriminant is 0), then z 1 ` z c1 c and z 1 z c 0 c. Proof (*). Let c z ` c 1 z ` c 0 0 be a quadratic equation. We know that its solutions are z 1 c 1 ` δ 1 and z 1 c 1 ` δ where δ 1 and δ are the two square roots of the discriminant. Moreover, we have δ ` δ 1 0. Therefore, and z 1 ` z c 1 ` δ 1 c 1 ` δ c1 c z 1 z p c 1 ` δ 1 qp c 1 ` δ q 4c c 1 δ 1 4c c 1 4c which concludes the proof. p c 1 ` δ 1 qp c 1 δ 1 q 4c 4c c 0 4c c 0 c

18 18 LAURENT DEMONET Example Thanks to Proposition 4.9, we can solve systems of equations of the form " z ` z 1 S zz 1 P where S, P P C and z and z 1 are indeterminate. Indeed, we know that z and z 1 are the solutions of the equation z Sz ` P 0. For example, if we want to solve " z ` z 1 1 i zz 1 i we have to solve z p 1 iqz ` i 0 and, using Example 4.8, we get that the solutions are " " z 1 z i z 1 and i z 1 1. You can easily check that these numbers satisfy the equations The fundamental theorem of algebra. Definition We call polynomial in C any function of the form: P pxq a n X n ` a n 1 X n 1 ` ` a 1 X ` a 0 where a 0, a 1,..., a n 1, a n are complex numbers such that a n 0. It is said to be constant if n 0. The degree of P is n. A root of P is a solution in C of the equation P pxq 0. Up to now, we saw that polynomials of the form X n a and a X `a 1 X ` a 0 have roots in C (see Propositions 4.3 and 4.7). In these cases, we even saw a method to compute these solutions. Now, we will see an important theorem which states that, in C, every non constant polynomial has at least one root. Unfortunately, there is no general way to compute these roots. Theorem 4.1 (Fundamental theorem of algebra). Let P be a non constant polynomial in C. Then it has at least one root. You should not learn this proof but you should try to understand the general strategy which will reappear later in the course. Proof. Let P be such a polynomial. The function z ÞÑ P pzq is continuous (see it as a function from R to R). Moreover, using (several times) Proposition 3.4, we have for z P C, P pzq ˇˇa n z n ` a n 1 z n 1 ` ` a 0ˇˇ a n z n ˇˇa n 1 z n 1ˇˇ ˇˇa n z n ˇˇ a 1 z a 0 ˆ z n a n a n 1 a n z z a 0 z n the second factor of which tends to a n when z tends to infinity. So, as a n 0, P pzq tends to `8 when z tends to infinity. In particular, there exists r such that, if z ą r then P pzq ą P p0q. From the course Calculus, we know that the function z ÞÑ P pzq admits a minimum m at a certain point z 0 on the disc of center 0 and radius r.

19 Moreover, if z is outside this disc, then COMPLEX NUMBERS 19 P pzq ą P p0q P pz 0 q m so in fact, m is the minimum of z ÞÑ P pzq on all C. Now, we can define a new polynomial Q by QpXq P px ` z 0 q such that it is clear that Q takes exactly the same values than P. In particular, z ÞÑ Qpzq reach its minimum m at 0 (because Qp0q P pz 0 q). If m 0, then Q has a root at 0 and P has a root at z 0. So we shall suppose that m ą 0. If we denote QpXq b 0 ` b 1 X ` ` b n X n we see that m Qp0q b 0. As we supposed that P is non constant, Q is also non constant. Therefore, there exists a smaller k 1 such that b k 0 and QpXq b 0 ` b k X k ` b k`1 X k`1 ` ` b n X n. Here comes the only part of the proof where we really use that we are in C and not in R. So this is the important part of the proof. Let z be an k-th root of b 0 {b k. Note that here, we do not try to compute it. Knowing that it exists is enough. If ε P r0, 1s, ˇ Qpεzq ˇb 0 ` b k ε k z k ` b k`1 ε k`1 z k`1 ` ` b n ε n z nˇˇˇ ˇ ˇb0 b k ε k b 0 ` b k`1 ε k`1 z k`1 ` ` b n ε n z n b k ˇ ˇ k ˇb 0 1 ε ` b k`1 ε k`1 z k`1 ` ` b n ε n z nˇˇˇ ˇ ˇb 0 1 ε k ˇˇˇ ` ˇˇˇbk`1 ε k`1 z k`1ˇˇˇ ` ` bn ε n z n (Prop. 3.4) ˇ b 0 ˇ1 ε kˇˇˇ ` ε k`1 ˇ ˇb k`1 z k`1ˇˇˇ ` ` ε n b n z n m 1 k ε ` ε ˇˇˇbk`1 k`1 z k`1ˇˇˇ ` ` ε n b n z n as 1 ε k 0 m m ε k ˇ ε ˇb k`1 z k`1ˇˇˇ ε n k b n z n. Now, notice that when ε tends to 0, ˇ m ε ˇb k`1 z k`1ˇˇˇ ε n k b n z n tends to m ą 0. In particular, we can find ε 0 P r0, 1s such that ε 0 ą 0 and m ε 0 ˇˇˇbk`1 z k`1ˇˇˇ ε n k 0 b n z n ą m. Thus Qpε 0 zq m ε k 0 m ε 0 ˇˇˇbk`1 z k`1ˇˇˇ ε n k 0 b n z n ă m ε k m 0 ă m. So we proved that Qpε 0 zq ă Qp0q even if Qp0q is a minimum of z ÞÑ Qpzq. It is a contradiction. More precisely, it contradicts the only hypothesis we did and we were not sure of, that is m ą 0. So m 0 and, as we saw before in this case, Q admits a root at 0 and P admits a root at z 0.

20 0 LAURENT DEMONET Remark Once again, pay attention that we can not compute in general a root of a polynomial even if we proved it admits one. In mathematics, we call such a proof non constructive because it does not construct the result but just prove that it exists. In fact, we can prove a little bit more. Namely, in a certain sense, the number of roots of a polynomial is exactly its degree if we consider that some roots are repeated (like double roots in the case of quadratic equations for example). Another way to express that is that we can always write a polynomial P in C in the form P pxq a n px z 1 qpx z q... px z n q where n is the degree of the polynomial, a n 0 and the z 1, z are the roots of the polynomial, where certain can be repeated. address: demonet@math.nagoya-u.ac.jp