9 Complex numbers and functions, factoring,and integration by parts9.1 Complex NumbersRecall that for every non-zero real number x, its square x2= x · x is alwayspositive. Consequently, R does not contain the square roots of any negativenumber. This is a serious problem which rears its head all over the place.It is a non-trivial fact, however, that any positive number has two squareroots in R, one positive and the other negative; the positive one is denoted√x. One can show that for any x in R,|x| =√x · x.So if we can somehow have at hand a square root of −1, we can find squareroots of any real number.This motivates us to declare a new entity, denoted i, to satisfyi2= −1.One defines the set of complex numbers to beC = {x + iy |x, y ∈ R}and defines the basic arithmetical operations in C as follows:(x + iy) ±(x′+ iy′) = (x ±x′) + i(y ± y′),and(x + iy)(x′+ iy′) = (xx′− yy′) + i(xy′+ x′y).There is a natural one-to-one functionR → C, x → x + i.0,compatible with the arithmetical operations on both sides.It is an easy exercise to check all the field axioms, except perhaps for theexistence of multiplicative inverses for non-zero complex numbers. To thisend one defines the complex conjugate of any z = x + iy in C to bez = x −iy .1Clearly,R = {z ∈ C |z = z}.If z = x + iy, we have by definition,zz = x2+ y2.In particular, zz is either 0 or a positive real number. Hence we can find anon-negative square root of zz in R. Define the absolute value, sometimescalled modulus or norm, by|z| =√zz =√x2+ y2.If z = x + iy is not 0, we will putz−1=zzz=xx2+ y2− iyx2+ y2.It is a complex number satisfyingz(z−1) = zzzz= 1.Done.It is natural to think of complex numbers z = x + iy as being orderedpairs (x, y) of real numbers. So one can try to visualize C as a plane with twoperpendicular coordinate directions, namely giving the x and y parts. Notein particular that 0 corresponds to the origin O = (0, 0), 1 to (1, 0) and i with(0, 1). Geometrically, one can think of getting from −1 to 1 (and back) byrotation about an angle π, and similarly, one gets from i to its square −1 byrotating by half that angle, namely π/2, in the counterclockwise direction.To get from the other square root of −1, namely −i, one rotates by π/2 inthe clockwise direction. (Going counterclockwise is considered to be in thepositive direction in Math.)Addition of complex numbers has then a simple geometric interpretation:If z = x+iy, z′= x′+iy′are two complex numbers, represented by the pointsP = (x, y) and Q = (x′, y′) on the plane, then one can join the origin O to Pand Q, and then draw a parallelogram with the line segments OP and OQas a pair of adjacent sides. If R is the fourth vertex of this parallelogram, itcorresponds to z + z′. This is called the parallelogram law.Complex conjugation corresponds to reflection about the x-axis.2The absolute value or modulus |z| of a complex number z = x + iy is, bythe Pythagorean theorem applied to the triangle with vertices O, P = (x, y)and R = (x, 0), simply the length, often denoted by r,√x2+ y2of the lineOP.The angle θ between the line segments OR and OP is called the argumentof z. The pair (r, θ) determines the complex number z. Indeed High schooltrigonometry allows us to show that the coordinates of z are given byx = rcosθ and y = rsinθ,where cos (or cosine) and sin (or sine) are the familiar trigonometric func-tions. Consequently,z = r(cos θ + isinθ).Those who know about exponentials (to be treated below in section 9.4) willrecognize the identityeiθ= cosθ + isinθ.(This can also be taken as a definition of eiθ, for any θ ∈ R.)Note that eiθhas absolute value 1 and hence lies on the unit circle in theplane given by the equation |z| = 1.It is customary for the angle θ to be called the argument of z, denotedarg(z), taken to lie in [0, 2π).9.2 Cardano’s formulaThis section is mainly for motivational purposes. Recall the well knownquadratic formula from the days of old, which asserts that the roots of thequadratic equationax2+ bx + c = 0, with a, b, c ∈ R,are given byα±=−b ±√D2a,where the discriminant D is b2− 4ac. Note thatD > 0 =⇒ ∃2 real roots;D = 0 =⇒ ∃ a unique real root (with multiplicity 2);3D < 0 =⇒  ∃ real root.There were several people in the old days (up to the middle of the nine-teenth century), some of them even quite well educated, who did not believein imaginary numbers, such as square-roots of negative numbers. Their re-action to the quadratic formula was to just ignore the case when D < 0and thus not deal with the possibility of non-real roots. They said they wereonly interested in real roots. Their argument was shattered when one startedlooking at the cubic equationax3+ bx2+ cx + d = 0, with a, b, c, d ∈ R.Thanks to a beautiful formula of the Italian mathematician Cardano, theroots are given byα1= S + T −b3a,α2= −(S + T)/2 −b3a+√−32(S − T ),α3= −(S + T)/2 −b3a−√−32(S − T ),withS = (R +√D)1/3, T = (R −√D)1/3,where the discriminant D is Q3+ R2, andR =9abc − 27a2d − 2b354a3, Q =3ac − b29a2.One hasD > 0 =⇒ ∃ a unique real root;D = 0 =⇒ ∃3 real roots with 2 of them equal;D < 0 =⇒ ∃3 distinct real roots.This presented a problem for the Naysayers. One is for sure interested inthe case when there are three real roots, but Cardano’s formula for the rootsgoes through an auxiliary computation, namely that of the square-root of D,which involves imaginary numbers!49.3 Complex sequences and seriesAs with real sequences, given a sequence {zn} of complex numbers zn, wesay that it converges to a limit L, say, in C iff we have, for every ε > 0, wecan find an integer N > 0 such thatn ≥ N =⇒ |L −zn| < ε.Proposition 1 (i) If {an} is a convergent sequence with limit L, then forany scalar c, the sequence {can} is convergent with limit cL;(ii) If {an}, {bn} are convergent sequences with respective limits L1, L2,then their sum {an+ bn} and their product {anbn} are convergent withrespective limits L1+ L2and L1L2.The proof is again a simple application of the properties of absolute val-ues. The following Corollary allows the convergence questions for complexsequences to b reduced to real ones.Corollary 9.1 Let {zn= xn+ iyn} be a sequence of complex numbers, withxn, ynreal for each n. Then {zn} converges iff the real sequences {xn} and{yn} are both convergent.Proof. Suppose {xn}, {yn} are both convergent, with