1. 1 Week 1 Complex numbers: the basics 1. The definition of complex numbers and basic operations 2. Roots, exponential function, and logarithm 3. Multivalued functions, or dependences

2. 2 where i ‘marks’ the second component. x and y are called the real and imaginary parts of z and are denoted 1. The definition of complex numbers and basic operations Example 1: ۞ The set of complex numbers can be viewed as the Euclidean vector space R 2, of ordered pairs of real numbers (x, y), written as Like any vectors, complex numbers can be added and multiplied by scalars. Calculate: (a) (1 + 3i) + (2 – 7i), (b) (–2)×(2 – 7i).

3. 3 ۞ Given a complex number z =x + i y, the +tive real expression is called the absolute value, or modulus of z. It’s similar to the absolute value (modulus, norm, length) of a Euclidean vector. Theorem 1: polar representation of complex numbers A complex number z = x + i y can be represented in the form where r = | z | and θ is the argument of z, or arg z, defined by

4. 4 Comment: arg z is measured in radians, not degrees! You can still use degrees for geometric illustrations. Like any 2D Euclidean vectors, complex numbers are in a 1-to-1 correspondence with points of a plane (called, in this case, the complex plane).

5. 5 Example 2: Show on the plane of complex z the sets of points such that: (a) | z | = 2, (b) arg z = π/3.

6. 6 Example 3: Show z = 1 + i on the complex plane and find θ = arg z. How many values of θ can you come up with? Thus, arg z is not a function, but a multivalued function, or a dependence. ۞ The principal value of the argument of a complex number z is denoted by Arg z (with a capital “ A ”), and is defined by Multivalued functions will be discussed in detail later. In the meantime, we introduce a single-valued version of arg z.

7. 7 Comment: The graph of arg z looks like a spiral staircase. Arg z

8. 8 For any z 1 and z 2, it holds that | z 1 + z 2 | ≤ | z 1 | + | z 2 |. Theorem 2: the Triangle Inequality Proof (by contradiction): Assume that Theorem 2 doesn’t hold, i.e. hence...

9. 9 Our strategy: get rid of the square roots – cancel as many terms as possible – hope you’ll end up with something clearly incorrect (hence, contradiction). The l.h.s. and r.h.s. of this inequality can be assumed +tive (why?) – hence, This inequality is clearly incorrect (why?) – hence, contradiction. █ Since the l.h.s. and r.h.s. of (1) are both +tive (why do we need this?), we can ‘square’ them and after some algebra obtain (1)

10. 10 ۞ The product of z 1 = x 1 + i y 1 and z 2 = x 2 + i y 2 is given by Example 4: In addition to the standard vector operations (addition and multiplication by a scalar), complex numbers can be multiplied, divided, and conjugated. Calculate (1 + 3i) (2 – 7i).

11. 11 or Example 5: Observe that One cannot, however, write because the square root is a multivalued function (more details to follow). Remark: When multiplying a number by itself, one can write z×z = z 2, z×z×z = z 3, etc.

12. 12 Example 6: ۞ The quotient z =z 1 /z 2, where z 2 ≠ 0, is a complex number such that Calculate (1 + 3i)/(2 – 7i).

13. 13 Theorem 3: multiplication of complex numbers in polar form Useful formulae: sin θ 1 cos θ 2 + sin θ 2 cos θ 1 = sin (θ 1 + θ 2 ), cos θ 1 cos θ 2 – sin θ 1 sin θ 2 = cos (θ 1 + θ 2 ). Proof: by direct calculation. where r 1,2 = | z 1,2 | and θ 1,2 = Arg z 1,2.

14. 14 Theorem 5: The de Moivre formula This theorem follows from Theorem 4 with | z | = 1. Theorem 4: where r = | z | and θ = Arg z. This theorem follows from Theorem 3.

15. 15 ۞ Complex numbers z 1 = x + i y and z 2 = x − i y are called complex conjugated (to each other) and are denoted by or Example 7: If z = 5 + 2i, then z* = 5 – 2i.

16. 16 Theorem 6: Proof: by direct calculation.

17. 17 2. Roots, the exponential function, and the logarithm ۞ The n th root of a complex number z is a complex number w such that The solutions of equations (2) are denoted by (2)

18. 18 where r = | z |, θ = Arg z, and k = 0, 1... n – 1. Theorem 7: This theorem follows from Theorem 4. For any z ≠ 0, equation (2) has precisely n solutions:

19. 19 Geometrical meaning of roots: To calculate z 1/2 where z = –4, draw the following table: | z | arg z | z 1/2 | arg z 1/2 = ½ (arg z) 4π 2 ½ π 4π + 2π2½ (π + 2π) Hence,

20. 20 Comment: You need to memorise the following values of sines and cosines: θsin θcos θ 001 π /61/2√3/2 π /4 √  √2/2 π /3√3/21/2 π /210 The symbol √ in the above table denotes square roots.

21. 21 Example 9: Find all roots of the equation w 3 = –8 and sketch on the complex plane. Example 8: Find: (a) sin 5π/4, (b) cos 2π/3, (c) sin (–5π/6).

22. 22 ۞ The complex exponential function is defined by Example 10: Find all z such that Im e z = 0. Comment: The polar representation of complex numbers can be re-written in the form where r = | z | and θ = arg z. (3)

23. 23 Comment: Consider and observe that any value of w corresponds to a single value of z. The opposite, however, isn’t true, as infinitely many values of w (differing from each other by multiples of 2πi ) correspond to the same value of z. This suggests that, even though the exponential is a single-valued function, the logarithm is not.

24. 24 ۞ The complex number w is said to be the natural logarithm of a complex number z, and is denoted by w = ln z, if (4) Theorem 8: This theorem follows from equalities (3)–(4). For any z ≠ 0, equation (4) has infinitely many solutions, such that where k = 0, ±1, ±2, ±3... (5) Example 11: Use (5) to calculate: (a) ln (–1), (b) ln (1 + i).

25. 25 ۞ The principle value of the logarithm is defined by ۞ General powers of a complex number z are defined by Since this definition involves the logarithm, z p is a multivalued function. It has, however, a single-valued version, Comment: Evidently, the logarithm, all roots, and non-integer powers of z are all multivalued functions.