Complex numbers from A to Z

[...]... + z1 + z2 z1 z2 = = A, 1 1 1 + z1 z2 1+ · z1 z2 12 1 Complex Numbers in Algebraic Form Problem 3 Let a be a positive real number and let M a = z ∈ C∗ : z + 1 =a z Find the minimum and maximum value of |z| when z ∈ Ma 1 Solution Squaring both sides of the equality a = z + , we get z a2 = z + 1 z 2 = z+ = 1 z z+ 1 z = |z| 2 + z 2 + (z) 2 1 + 2 |z| 2 |z| |z| 4 + (z + z) 2 − 2 |z| 2 + 1 |z| 2 Hence |z| 4 − |z| 2... moduli) z2 |z 2 | 9) |z 1 | − |z 2 | ≤ |z 1 − z 2 | ≤ |z 1 | + |z 2 | Proof One can easily check that (1)–(4) hold (5) We have |z 1 · z 2 |2 = (z 1 · z 2 ) (z 1 · z 2 ) = (z 1 · z 1 ) (z 2 · z 2 ) = |z 1 |2 · |z 2 |2 and consequently |z 1 · z 2 | = |z 1 | · |z 2 |, since |z| ≥ 0 for all z ∈ C (6) Observe that |z 1 + z 2 |2 = (z 1 + z 2 ) (z 1 + z 2 ) = (z 1 + z 2 ) (z 1 + z 2 ) = |z 1 |2 + z 1 · z 2 + z 1 · z. .. equation with |z 1 | = 1 From z 2 = c 1 · a z1 c b 1 · = 1 Because z 1 + z 2 = − and |a| = |b|, we have a |z 1 | a |z 1 + z 2 |2 = 1 This is equivalent to it follows that |z 2 | = (z 1 + z 2 ) (z 1 + z 2 ) = 1, i.e., (z 1 + z 2 ) We find that (z 1 + z 2 )2 = z 1 z 2 , i.e., − b a 1 1 + z1 z2 2 = = 1 c , a which reduces to b2 = ac, as desired b) As we have already seen, we have b2 = ac and c2 = ab Multiplying... + |z 2 |2 Because z 1 · z 2 = z 1 · z 2 = z 1 · z 2 it follows that z 1 z 2 + z 1 · z 2 = 2 Re (z 1 · z 2 ) ≤ 2 |z 1 · z 2 | = 2 |z 1 | · |z 2 |, hence |z 1 + z 2 |2 ≤ ( |z 1 | + |z 2 |)2 , and consequently, |z 1 + z 2 | ≤ |z 1 | + |z 2 |, as desired In order to obtain inequality on the left-hand side note that |z 1 | = |z 1 + z 2 + ( z 2 )| ≤ |z 1 + z 2 | + | − z 2 | = |z 1 + z 2 | + |z 2 |, hence |z. .. − |z 2 | ≤ |z 1 + z 2 | (7) Note that the relation z · |z −1 | = |z| −1 (8) We have 1 1 1 1 = 1 implies |z| · = 1, or = Hence z z z |z| z1 1 |z 1 | −1 −1 = z1 · = |z 1 · z 2 | = |z 1 | · |z 2 | = |z 1 | · |z 2 |−1 = z2 z2 |z 2 | 1.1 Algebraic Representation of Complex Numbers 11 (9) We can write |z 1 | = |z 1 − z 2 + z 2 | ≤ |z 1 − z 2 | + |z 2 |, so |z 1 − z 2 | ≥ |z 1 | − |z 2 | On the other hand,... multiplication The multiplication of complex numbers satisfies the following properties: (a) Commutative law z 1 · z 2 = z 2 · z 1 for all z 1 , z 2 ∈ C (b) Associative law (z 1 · z 2 ) · z 3 = z 1 · (z 2 · z 3 ) for all z 1 , z 2 , z 3 ∈ C (c) Multiplicative identity There is a unique complex number 1 = (1, 0) ∈ C such that z · 1 = 1 · z = z for all z ∈ C A simple algebraic manipulation is all that is... Because z · 1 1 = 1, we have z · z z = 1, and consequently z · (z −1 ) = (z) −1 z1 1 = z1 · z2 z2 8) From the relations 7) Observe that = z1 · 1 z2 = z1 · z1 1 = z2 z2 z + z = (x + yi) + (x − yi) = 2x, 1 z = 1, yielding 1.1 Algebraic Representation of Complex Numbers 9 z − z = (x + yi) − (x − yi) = 2yi it follows that Re (z) = z+ z 2 and Im (z) = z z 2i as desired The properties 4) and 5) can be easily... + |z 2 |2 + |z 1 |2 − z 1 · z 2 − z 2 · z 1 + |z 2 |2 = 2( |z 1 |2 + |z 2 |2 ) Problem 2 Prove that if |z 1 | = |z 2 | = 1 and z 1 z 2 = −1, then number z1 + z2 is a real 1 + z1 z2 Solution Using again property 4 in the above proposition, we have z 1 · z 1 = |z 1 |2 = 1 and z 1 = Likewise, z 2 = 1 z1 1 Hence denoting by A the number in the problem we have z2 z1 + z2 A= = 1 + z1 · z2 so A is a real... identities: a) |z 1 + z 2 |2 + |z 2 + z 3 |2 + |z 3 + z 1 |2 = |z 1 |2 + |z 2 |2 + |z 3 |2 + |z 1 + z 2 + z 3 |2 ; b) |1 + z 1 z 2 |2 + |z 1 − z 2 |2 = (1 + |z 1 |2 )(1 + |z 2 |2 ); c) |1 − z 1 z 2 |2 − |z 1 − z 2 |2 = (1 − |z 1 |2 )(1 − |z 2 |2 ); d) |z 1 + z 2 + z 3 |2 + | − z 1 + z 2 + z 3 |2 + |z 1 − z 2 + z 3 |2 + |z 1 + z 2 − z 3 |2 = 4( |z 1 |2 + |z 2 |2 + |z 3 |2 ) 16 Let z ∈ C∗ such that z 3 + 1... 25 An integer power of a complex number z ∈ C∗ is defined by z 0 = 1; z 1 = z; z 2 = z · z; z n = z · z · · · z for all integers n > 0 n times (z −1 )−n and = for all integers n < 0 The following properties hold for all complex numbers z, z 1 , z 2 ∈ C∗ and for all integers m, n: 1) z m · z n = z m+n ; zm 2) n = z m−n ; z 3) (z m )n = z mn ; n n 4) (z 1 · z 2 )n = z 1 · z 2 ; n n z z1 5) = 1 n z2 z2 . several international journals. Dorin has been a regular faculty member at the Canada–USA Mathcamps since 2001. Titu Andreescu Dorin Andrica Complex Numbers fromAto Z Birkh ¨ auser Boston • Basel • Berlin Titu. Analysis and Applications.” Professor Andreescu currently teaches at the University of Texas at Dallas. Titu is past chairman of the USA Mathematical Olympiad, served as director of the MAA American. faculty member at the Canada–USA Mathcamps since 2001. Titu Andreescu Dorin Andrica Complex Numbers fromAto Z Birkh ¨ auser Boston • Basel • Berlin Titu Andreescu University of Texas at Dallas School