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INDIAN INSTITUTE OF TECHNOLOGYBOMBAY

MA 204 - Mathematics IVLecture 1

Basic Properties of Complex Numbers

1 Prerequisites1.1 Reals Numbers:I The law of commutativity: a + b = b + a; ab = ba, for all a, b R.II The law of associativity: (a + b) + c =

a + (b + c); (ab)c = a(bc), for all a,b,c R.III The law of distributivity: (a + b)c = ac + bc, for all a,b,c R.IV The law of identity: a + 0 = a; a1 = a, for all a R.V The law of additive inverse: Given any a R, there exists a unique x R such thata + x = 0.

VI The law of multiplicative inverse: Given a R, a = 0, there exists a unique x Rsuch that ax = 1.

Furthermore, there is a total ordering 0.

1 Algebra of Complex Numbers

We define the algebra of complex numbers C to be the set of formal symbols x +

y, x, y R together with the addition and multiplication defined as follows:

(x1 + y1) + (x2 + y2) = (x1 + x2) + (y1 + y2);

(x1 + y1)(x2 + y2) = (x1x2

y1y2) + (x1y2 + y1x2).

2 + 1 = 0; i.e., 2 = 1.

Theorem 1 The setC of all formal expressions a + b where a, b R forms thesmallest field containingR as a subfield and in which is a solution of the equation

X2 + 1 = 0.

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Observe that a complex number is well-determined by the two real numbers, x, y

viz., z := x + y. These are respectively called the real part and imaginary part of z. We

write:

z = x; z = y. (1)

If

(z) = 0, we say z is (purely) imaginaryand similarly if

(z) = 0, then we say z

is real. The only complex number which is both real and purely imaginary is 0. Observe

that, according to our definition, every real number is also a complex number.

equating the real and the imaginary parts of the two sides of an

equation

is indeed a part of the definition of complex numbers and will play a very important

role.

2 Conjugation and Absolute Value

Definition 1 Following common practice, for z = x + y we denote by z = x y andcall it the (complex) conjugate of z. and call it

the conjugate of z.

(z) =

z + z

2

;

(z) =

z z2

. (2)

z1 + z2 = z1 + z2, z1z2 = z1 z2, z = z. (3)

Definition 2 Given z C, z = a + b, we define its absolute value (length ) |z| to bethe non-negative square root of a2 + b2, i.e.,

|z| := (a2 + b2).

Remark 1 |z|2 = zz. Thereforez C, |z| = 0 z = 0.Also, for z = 0,

z1 = z|z|2.

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Basic Identities and Inequalities

(B1) |z| = |z|.(B2) |z1z2| = |z1||z2|.(B3) |(z)| |z| ( resp. |(z)| |z|); equality holds iff(z) = 0 (resp. (z) = 0).(B4) Cosine Rule:

|z1 + z2|2 = |z1|2 + |z2|2 + 2(z1z2).(B5) Parallelogram Law :

|z1 + z2|2 + |z1 z2|2 = 2(|z1|2 + |z2|2).

(B6) Triangle inequality : |z1 + z2| |z1| + |z2| and equality holds iff one of the zj is anon-negative multiple of the other.

(B7) Cauchys1 Inequality :

n

j=1

zj wj

2

nj=1

|zj|2

nj=1

|wj|2 .

Polar form:

0

r

r(cos, sin )

)(E

Fig. 0

Given (x, y) = z = 0, the angle , measured in counter-clockwise sense, made bythe line segment [0, z] with the positive real axis is called the argument or amplitude of

z :

= arg z.

1Augustin Louis Cauchy (1789-1857) was a French mathematician, an engineer by training. He did

pioneering work in analysis and the theory of permutation groups, infinite series, differential equations,

determinants, probability and mathematical physics.

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x = r cos ; y = r sin (4)

Let us temporarily set-up the notation

E() := cos + sin . (5)

Then the complex number z = x + y takes the form

z = r(cos + sin ) =: rE().

Observe |z| = r. Now let z1 = r1E(1), z2 = r2E(2). Using additive identities for sineand cosine viz.,

sin(1 + 2) = sin 1 cos 2 + cos 1 sin 2,

cos(1 + 2) = cos 1 cos 2 sin 1 sin 2,(6)

we obtain

z1z2 = r1r2E(1 + 2). (7)

If we further remind ourselves that the argument can take values (in radians)

between 0 and 2, then the above identity tells us that arg(z1z2) = arg z1 + arg z2

(mod 2) provided z1 = 0, z2 = 0.Put zj = rjE(j) for j = 1, 2, and let be the angle between the vectors represented

by these points. Then z1z2 = r1r2E(1 2) and hence (z1z2) = r1r2 cos . Thus,

cos =(z1z2)|z1z2| . (8)

Now, we can rewrite the cosine rule as:

|z1 + z2|2 = r21 + r22 + 2r1r2 cos . (9)

Note that by putting = /2 in (9), we get Pythagoras theorem.

Remark 2 Observe that given z = 0, arg z is a multi-valued function. Indeed, if isone such value then all other values are given by + 2n, where n Z. Thus to beprecise, we have

arg z = { + 2n : n Z}This is the first natural example of a multi-valued function. We shall come

across many multi-valued functions in complex analysis, all due to this nature of arg z.

However, while carrying out arithmetic operations we must select a suitable value for

arg from this set. One of these values of arg z which satisfies < arg z is singledout and is called the principal value of arg z and is denoted by Arg z.

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Example 1 The three cube roots of unity are

1, cos2

3+ sin

2

3, cos

4

3+ sin

4

3

which we can simplify as:

1,1 + 3

2,1 3

2.

Remark 3 deMoivres2Law Now observe that, by putting r1 = r2 = 1 in (7) we

obtain:

E(1 + 2) = E(1)E(2).

Putting 1 = 2 = and applying the above result repeatedly, we obtain

E(n) = E()n.

This is restated in the following:

deMoivres Law:

cos n + sin n = (cos + sin )n. (10)

Remark 4 Roots of complex numbers: Thanks to our geometric understanding, we can

now show that the equation

Xn = z (11)

has exactlyn roots inC for every non zero zC. Suppose w is a complex number that

satisfies the equation (in place of X,) we merely write

z = rE(Argz), w = sE(Arg w).

Then we have,

snE(nArg w) = wn = z = rE(Argz)

Therefore we must have s = n

r = n

|z| and arg w will contain the values

Arg z

n+

2k

n, k = 0, 1, . . . , n

1.

Thus we see that (11) has n distinct solutions. One of these values viz., n

|z|E( Arg zn

) is

called the principal value of the nth root function and is denoted by n

z.

2Abraham deMoivre(1667-1754) was a French mathematician. He also worked in Probability theory.

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INDIAN INSTITUTE OF TECHNOLOGYBOMBAY

Spring 2007

MA 204 - Mathematics IVLecture 2

/5th Jan 07

1 Geometry of Complex Numbers

The picture below illustrates how to add two complex numbers geometrically.

O

Q

P2

P1

x x1 2

2y

1y

y21+

x1+ x2

y

Fig.1

The parallelogram law (B5) now becomes:

The sum of the squares of the lengths of the diagonals of a

parallelogram is equal to the sum of the squares of the lengths

of the sides.

Remark 1 Geometric Multiplication of complex numbers:

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0

ZZ

1

2

1

Z1Z2

Fig. 2

In the picture above, the two triagles a similar. It tells you how to multiply two

complex numbers.

Similarly, to get the inverse of z2 = 0, we first re-scale the vector

OP2 to have

length r1

2

and then change its direction to have an amplitude 2. Also, it follows that

the amplitude of z1z12 is the angle that

OP1 makes with

OP2 .

It follows that given 0 = C, the assignment z z defines a linear map

R2 R

2, which is a composite of a rotation (through an angle = Arg ) and a

dilation or a scaling ( by a factor r = ||). Such linear maps R2 R

2are called

similarities. For the converse part you will have to wait.

The dot product Think of two complex numbers as vectors in R2. Then their dot

product is given by

(x1 + y1) (x2 + y2) = x1x2 + y1y2

which we can rewrite in terms of multiplication of complex numbers as:

z1 z2 = (z1z2) = (z1z2.

Similarly, very that that cross product can be rewritten is the form

z1 z2 = (z1z2)

Equation of a line: Let ax + by + c = 0 represent a line in cartesian coordinates.

WLOG we may assume that a2 + b=1. Put w = a + b; z = x + y. Then ax + by =

Re(wz) = 1/2(wz + wz). Thus, we see that the general equation of a line in the planecan be given by complex number as:

wz + wz = t, t R.

Equation of a circle:

(z w)(z w) = r2, r R.

Definition 1 By a rigid motion or an isometry of the plane, we mean a mapping

f : C C which preserves distances, i.e., |f(z) f(w)| = |z w| for all z, w C.

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Remark 2 Identity, translation, rotation about a point and reflexion in a line are ob-

vious examples of rigid motions. The first three preserve the sense of orientation of the

plane and the third one changes it. The first one keeps all points fixed. The second one

has no fixed point but it preserves orientation. The third one fixes exactly one point and

also preserves the oriention. The fourth one fixes exactly a line and (hence) changes the

orientation. Are there other rigid motions such as those which fix no points and change

the orientation? We shall investigate this right now.

It is clear that composite of two rigid motions is again a rigid motion. It is alsoclear that a rigid motion is a one-one mapping but it is not easy to see that it also onto.

This will follow as an easy consequence of what we are going to do.

Reflection in a Line Let

wz + wz = t

represent a line. If z denotes the image of z under the reflection in L, then z z is

parallel to w and is bisected by L. Therefore, we obtain,

z

z = sw,s R; & w(z + z) + w(z

+ z) = 2t.

Upon simplification this yields

z =t wz

w= wt w2z.

Definition 2 By a glide-reflection we mean a RM which is a reflection in a line followed

by a translation by a non zero vector in the direction of L.

It is easy to see that a glide-reflection does not have any fixed point and does not

preserve the orientation. The converse follows from what we see below.

Theorem 1 Let f : C C be a rigid motion such that f(0) = 0. Then there exist

unique a, b C with |a| = 1 such that

f(z) = az + b,z C OR f(z) = az + b,z C.Proof: Put b = f(0) and define g(z) = f(z) + b. Then g is also a RM. and g(0) = 0.

Now |g(1)| = 1. So, put a = g(1)1 and define h(z) = ag(z). Then h is RM and

h(0) = 0, h(1) = 1, and h() = .

Case 1: Assume h() = . Now consider any z = x + C and put h(z) = u + v. Then

it follows that

u2 + v2 = x2 + y2;

(u 1)2 + v2 = (x 1)2 = y2;

u2 + (v 1)2 = x2 + (y 1)2. Solving these, yields, u = x, v = y. Thus h(z) = z, z C.

This is the same as saying f(z) = az + b,z C.

Case 2: Assume that h() = . Put h(z) = h(z). Then h is a RM and h(0) = 0, h(1) =

1, h() = . So, we are in case 1. .

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Theorem 2 Letf : C C be a rigid motion.

(i) Suppose f fixes two distinct points. Then all points on the line passing through these

two points are also fixed by f.

(ii) Suppose f fixes three non collinear points. Then f = Id.

(iii) Suppose f fixes an entire line L. Then it is either Id or the reflexion in that line.

(iv) Suppose f fixes exactly one point. Then it is a rotation around that point.

(v) Suppose f fixes no points. Then either f is a translation or a glide reflexion.

Proof: (i) If z1, z2 are such that azi + b = zi, i = 1, 2 then for any real numbers t1, t2

such that t1 + t2 = 1, we have

a(t1z1 + t2z2) + b = t1z1 + t2z2.

(ii) Argue as in (i).

(iii) We may assume that 0, 1 are fixed by f, so that the entire real line is fixed. Now if

f fixes also, then by (ii), f = Id. If f() = then we consider the RM f which must

be Id.(iv) Choose the fixed point f as the origin. Then f(0) = 0. This means b = 0. Also,

observe that if f(z) = ez then f(re/2) = re), shows that f fixes a whole line.

Therefore f(z) = e)z for some .

(v) First consider the case f(z) = az + b. If a = 1 then we can solve for f(z) = z and

therefore a = 1. Now consider f(z) = az + b. If a = 1, it follows that f is a reflection

in the line y = b followed by translation by (b) = 0. Now say a = 1 Put 2 = a. If

b = t for some t R again f is the reflection in the line joining 0 and . So, b = t.

Therefore there exists s R such that s(1 a) = b t. Let L be the line parallel to

the vector and passint through s. Then it follows that f is the reflection in the line Lfollowed by the translation by the vector t.

Story Time This story is due to George Gamow, the well-known physicist and

an ingenious story-teller. We quote from his book:

ONE TWO THREE INFINITY (pp. 44-45).

There was a young and adventurous man who found among his great-grand fa-

thers papers a piece of parchment that revealed the location of a hidden treasure. The

instructions read:

Sail to North latitude and West longitude where thou wilt find a deserted

island. There lieth a large medow, not pent, on the north shore of the island where

standeth a lonely oak and a lonely pine. There thou wilt see also an old gallows on

which we once were wont to hang traitors. Start thou from the gallows and walk to the

oak counting thy steps. At the oak thou must turn right by a right angle and take the

same number of steps. Put here a spike in the ground. Now must thou return to the

gallows and walk to the pine counting thy steps. At the pine thou must turn left by a

right angle and see that thou takest the same number of steps, and put another spike

into the ground. Dig half-way between the spikes; the treasure is there.

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Lecture 3

Complex DifferentiabiltyNotation: Br(z) = {w C : |w z| < r}

Open ball of radius r center z.

Definition 1 A C is said to be an open set if it is the union of open balls.

Thus if A is open and z A then it follows that there is r > 0 such that Br(z) A.

Definition 2 Letz0 A C, such that there exists r > 0 with Br(z0) A. Letf : A C

be a map. Then f is said to be complex differentiable (written Cdifferentiable at z0)

if the limit on the right hand side of the following formula exists, and in that case we call this

limit, the derivative of f at z0 :

df

dz(z0) := lim

h0

f(z0 + h) f(z0)

h. (1)

We also use the notation f(z0) for this limit and call it Cauchy derivative of f at

z0. For an open subset A ofC, if f is complex differentiable at each z0 A then we say f is

holomorphic on A. If f is holomorphic on A, the map z f(z) is called the derivative of f

on A and is denoted by f.

Remark 1 We caution you that the words holomorphic and analytic have been used bydifferent authors to mean somewhat different things. Luckily, this is not a serious matter, since

we shall see that they ultimately they all mean the same thing.

Example 1 Let us work out the derivative of the function f(z) = zn, where n is an integer,

directly from the definition. Of course, for n = 0, the function is a constant and hence, it is

easily seen that it is differentiable everywhere and the derivative vanishes identically. Consider

the case when n is a positive integer. Then by binomial expansion, we have,

f(z + h) f(z) = h

n1

zn1 +

n2

hzn2 + + hn1

.

Therefore, we have,

limh0

f(z + h) f(z)

h= nzn1.

This is valid for all values of z. Hence f is differentiable in the whole plane and its derivative

is given by f(z) = nzn1. Next, consider the case when n is a negative integer. We see that the

function is not defined at the pointz = 0. Hence we consider only points z = 0. Writingn = m

and f(z + h) f(z) =zm (z + h)m

(z + h)m

zm

and applying binomial expansion for the numerator as

above, we again see that

f(z) = m

zm+1= nzn1, z = 0.

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Remark 2 As in the case of calculus of one real variable, the Cauchy derivative has all the

standard properties: sum of two Cdifferentiable function is Cdifferentiable and so

on.

(f + g)(z0) = f(z0) + g

(z0), , C. (2)

Thus for each fixed z0, the assignment f f(z0) defines a linear functional on the space of all

complex differentiable functions defined in a neighborhood of z0. Moreover, just like in the real

case, it is also a derivation, i.e.,

(f g)(z0) = f(z0)g(z0) + f(z0)g

(z0). (3)

We even have the same formula for the derivative of a quotient:f

g

(z0) =f(z0)g(z0) f(z0)g

(z0)

(g(z0))2; if g(z0) = 0. (4)

The proof of the following theorem is exactly the same as the proof of the corresponding result

for real valued function of a real variable.

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Theorem 1 (The Increment Theorem:)

Letf : A C, z0 A, r > 0 such that Br(z0) A. Then f is complex differentiable at

z0 iff C, and a set theoretic function : Bs(0) \ {0} C, (0 < s < r) such that for all

h Bs(0) \ {0}

(f(z0 + h) f(z0) = h + h(h); limh0 (h) = 0. (5)

Proof: For the given f, we simply take

(h) :=f(z0 + h) f(z0)

h, h Br(0) \ {0}.

Then f is holomorphic at z0 iff limh0 (h) exists. In that case, we simply put equal to

this limit and take (h) = (h) and observe that (h) 0 iff (h) . On the otherhand, if there is such a function and a constant then clearly, limh0 (h) = , and so, f is

holomorphic at z0 and f(z0) = .

Remark 3 We may assume that the error function in (5) is defined on the whole of Br(0),

its value at0 being completely irrelevant for us. The increment theorem enables one to deal with

many tricky situations while dealing with differentiability. As an illustration we shall derive the

chain rule for differentiation.

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Theorem 2 (Chain Rule :) Let f : A C, g : B C, f(A) B and z0 A. Suppose

that f(z0) and g(f(z0)) exist. Then (g f)

(z0) exists and (g f)(z0) = g

(f(z0))f(z0).

Proof: Let

f(z0 + h) f(z0) = hf(z0) + h(h);

g(f(z0) + k) g(f(z0)) = kg(f(z0)) + k(k) ;

(6)

be as in the increment theorem, where

(h) 0 as h 0; & (k) 0 as k 0

Let us say : Bs(0) C and : Br(0) C. Now, if we take s to be sufficiently small, then

it follows (from the continuity of f at z0) that for h Bs(0), we have f(z0 + h) f(z0) Br(0).

Hence we can put k = f(z0 + h) f(z0), in (6) to obtain,

g(f(z0 + h)) g(f(z0)) = [hf(z0) + h(h)][g

(f(z0)) + (k)]

= hg(f(z0))f(z0) + h(h),

where, (h) = (h)g(f(z0)) + ((h) + f(z0))(f(z0 + h) f(z0)). Observe that as h 0, we

have, k = f(z0 + h) f(z0) 0 and (k) 0. Hence (h) 0, as h 0. Thus by the

increment theorem again, (g f)(z0) exists and is equal to g(f(z0))f

(z0) as desired.

Exercise: (i) Derive the 00

-form of LHospital rule for complex differentiable functions. (Hint:Use

increment theorem.) Apply this to find the following limit:

limzz0

1

z z0

1

( z)n

1

( z0)n

.

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The following result, which is completely obvious from the definition of differentiability,

is extremely useful in extending the domain of definition of a holomorphic function.

Theorem 3 Patching-up Lemma: LetU1, U2 be any two open sets inC and fi : U C be

holomorphic functions. If f1(z) = f2(z) for all z U1 U2 then the function defined by

g(z) =

f1(z) z U1f2(z) z U2

is holomorphic on U1 U2.

Remark 4 So far, we have considered derivatives of functions at a point z A only when the

function is defined in a nbd of z. We can try to relax this condition as follows: Thus, if B C

and f : B C, we say f is holomorphic on B if f extends to a holomorphic map f : A C

where A is an open subset containing B. However, we can no longer attach a unique derivative

to f at points of B in general. With some more suitable geometric assumptions on B, this can

be made possible. For instance if B is a closed disc or a closed rectangle, and if f : B C is

differentiable on B, then even at the boundary points of B, the derivative of f is unique.

Cauchy-Riemann Equations

Let U be an open subset ofC, z0 = (x0, y0) U and f : U C be a given function.

Write z = x + y, z0 = x0 + y0, h = h1 + h2, f(x + y) = u(x, y) + v(x, y). Supposef(z0) exists and let f

(z0) = + . By the increment theorem, we have

f(z0 + h) f(z0) = hf(z0) + h(h); (h) 0 as h 0. (7)

[Put L(h) = hf(z0). Then L : C C is a real linear map. Also

|f(z0 + h) f(z0) L(h)|

|h| 0; as h 0.

Therefore, it follows that f is real (total) differentiable at z0 and D(f)(z0) = L. In particular,

all directional derivatives of f exist at z0.]

Put (h) = 1(h) + 2(h), where 1, 2 are real and imaginary parts of . Then clearly,

j(h) 0 as h 0. (8)

Put h = h1 in (7), i.e., take h2 = 0, to obtain

f(x0 + h1, y0) f(x0, y0) = h1( + ) + h1(h1). (9)

Comparing the real and the imaginary parts on both the sides we get

u(x0 + h1, y0) u(x0, y0) = h1 + h11(h1);

and v(x0 + h1, y0) v(x0, y0) = h1+ h12(h1).

(10)

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Therefore, by the increment theorem for 1-variable functions, it follows that ux and vx

exist at (x0, y0) and we have,

ux(x0, y0) = , vx(x0, y0) = . (11)

Now put h = h2, i.e., take h1 = 0 in (7), to obtain

f(x0, y0 + h2) f(x0, y0) = h2( + ) + h2(h2)

Again, comparing the real and imaginary parts and using increment theorem, we obtain

uy(x0, y0) = ; vy(x0, y0) = . (12)

Thus (11) and (12) together give

ux(x0, y0) = vy(x0, y0); uy(x0, y0) = vx(x0, y0) (13)

These are called CauchyRiemann(CR)-equations.1 Observe that we also have.

f(z0) = ux(x0, y0) + vx(x0, y0) (14)

= vy(x0, y0) uy(x0, y0) (15)

and

|f

(z0)|

2

= u

2

x + v

2

x = u

2

y + v

2

y = u

2

x + u

2

y = v

2

x + v

2

y = uxvy uyvx. (16)

The last expression above, which is the determinant of the matrix

ux uy

vx vy

(17)

is called the jacobian of the mapping f = (u, v), with respect to the variables (x, y) and is

denoted by

J(x,y)(u, v) := uxvy uyvx. (18)

1Bernhard Riemann(1826-1866) a German mathematician.

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Remark 5 A simple minded application of CR-equations is that it helps us to detect easily when

a function is not C-differentiable. For example, (z), (z) etc are not complex differentiable

anywhere. The function z |z|2 is not complex differentiable for any point z = 0. However,

it satisfies the CR-equations at 0. That of course does not mean that it is C-differentiable at 0.

(See the exercise below.)

Example 2 If f(z) = z2

z, z = 0 and f(0) = 0, show that Cauchy-Riemann equations are

satisfied at z = 0, but f(0) does not exist.

Sol:

For the first part, put f = u + v. Then u(x, y) = x33xy2

x2+y2; v(x, y) = x

2y+y3

x2+y2. Direct

computation shows that ux(0, 0) = 1; uy(0, 0) = 0; vx(0, 0) = 0, vy(0, 0) = 1. Hence CR equations

are satisfied.

However, for z = rei, we have

limr0

f(z) f(0)

z= lim

r0

z2

z= e3i.

which shows that f(0) = limz0z2

zdoes not exist.

Cauchy-Riemann equations under orthogonal Transformations

If f(z) = u + iv is differentiable at a point z0 = z0 + iy0 of a domain G, then

u

s=

v

n,

u

n=

v

s()

at (x0, y0) wheres

and n

denote directional differentiation in two orthogonal directions s and

n at (x0, y0), such that n is obtained from s by making a counterclockwise rotation.

Solution: Put s = cos i + sin j, n = sin i + cos j. Then

us

=u

s= u s = ux cos + uy sin ;

vs

=v

s= v s = vx cos + yy sin .

Similarly find un, vn. These quantities can be expressed in matrix form as:

us un

vs vn

=

ux uy

vx vy

cos sin

sin cos

=

ux uy

vx vy

R

R is rotation through an angle which is a multiplication by a non zero complex number.

Therefore the first matrix is a multiplication by a complex number iff the second one is. In

view of the above ex. this establishes (*).

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CR equations in polar coordinatesTake a point other than the origin. (At the origin polar coordinate is singular.) Say,

z0 = (x0, y0) = (0, 0) and let f = u + v. We claim that equations

rur = v; rvr = u. (19)

are equivalent to CR-equations and obtain the formula:

f(z0) = ei0(ur + ivr) =

iz0

(u + iv) (20)

Sol: Since x = r cos ; y = r sin , by chain rule:

xr = cos ; x = r sin ; yr = sin ; y = r cos .Moreover, ur = ux cos + uy sin ; vr = vx cos + vy sin ;

u = uxr sin + uyr cos ; v = vxr sin + vyr cos .

The latter can be expressed in the matrix form as:

r

cos sin

sin cos

ux vy

uy vx

=

rur v

u rvr

.

Rewriting the above matrix equation in the form

ux vyuy vx

= cos sin sin cos

ur v/ru/r vr

and substituting in f(z) = ux + uy, gives

f(z) = cos ur + sin ur

+ i

sin ur + cos ur

= cos ur + sin vr i (sin ur + cos vr)

= ei(ur + ivr)

=ei

r

(v iu) =i

z

(u + iv).

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Lecture 4

Theorem 1 If f is differentiable at a point then it is continuous at that point.

Theorem 2 Let U be an open subset of C and f : U R be a function. Then f is real-

differentiable on U and the function f : U C is continuous iff both the partial derivatives of

f exist on U and are continuous on U.

We shall not give a proof of this theorem here. It suffices to remark that one has to use

the mean value theorem of 1-variable calculus. Such a function is said to be continuously

differentiable at x.

Observe that the assignment x (Df)x

defines a map Df of U into the space of

all linear maps R2 into R viz., the dual vector space R2 which is isomorphic to R2. So, one

can define f to be continuously differentiable if Df is defined and continuous. Since the two

coordinate functions of Df are nothing but the two partial derivatives, the continuity of Df isequivalent to that of the continuity of the two partial derivatives of f.

Real-linear VS Complex Linear

Let us first consider a purely algebraic problem: Treating C as a 2-dimensional real vector

space, consider a real linear map T : C C given by the matrix a b

c d

When is it a complex linear map? We see that, if T is complex linear, then T() = T(1) and

hence, b + d = T() = T(1) = (a + c). Therefore, b = c and a = d.

Conversely, a R-linear map T : C C given by a b

b a

has the property that

T(z) = T(x + iy) = ax + ibx + (by + iay) = (a + ib)(x + iy) = wz

which is merely multiplication by a complex number.

Theorem 3 Letf : U C be a continuous function, f = u + v, z0 = x0 + y0 be a point of U.

Then f is C-differentiable at z0 iff considered as a vector valued function of two real variables,

f is real differentiable at z0 and its derivative (Df)z0 : C C is a complex linear map. In

that case, we also have f(z0) = (Df)z0.

Proof: Suppose that f is C-differentiable at z0. Then we have already seen that f is real

differentiable and its total real derivative D(f)(z0) is equal to f(z0).

Conversely, assume that Df(z0) exists and is complex linear. This means that there exist

error functions and say, such that

u(z0 + h) u(z0) = (h1ux + h2uy) + |h|(h)

v(z0 + h) v(z0) = (h1vx + h2vy) + |h|(h)

(1)

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with (h) 0 and (h) 0 as h 0. Complex linearity of Df means that ux = vy :=

say, uy = vx := say. Then, multiply the second equation by and add it to the first

equation above to obtain

f(z0 + h) f(z0) = h( + ) + |h|((h) + (h)),

where,

limh0

((h) + (h)) = 0.

Finally, in this case, the complex linear map D(f)z0 given by the matrix (??) is nothing but

multiplication by the complex number ux + vx = f(z0). Hence, the proof of the theorem is

complete.

As an immediate corollary, we have,

Theorem 4 Let f : U C be a complex differentiable function in a convex domain U suchthat f(z) 0. Then f is a constant on U.

There are certain useful partial results that relate the two notions of differentiability. We

shall mention some of them here without proof. The most popular one is:

Theorem 5 Letf be a complex valued function of a complex variable defined on an open subset

U. Then f is complex differentiable in U iff both the first order partial derivatives of f exist,

are continuous and satisfy the CauchyRiemann equations in U.

Remark 1 In the if part of this theorem, the continuity of the partial derivatives ensures via,

MVT, the existence of real derivative. Now the Cauchy-Riemann equations are equivalent to

the complex linearity of the real derivative. Hence we can appeal to the previous theorem. In the

only if part, the only thing that is not proved already is the continuity of the partial derivatives.

We shall not prove it here. It will follow once we show that any Cauchy differentiable function

has a continuous derivative. Indeed, we shall see later on that complex differentiability ensures

that the function has continuous derivatives of all orders!

(Looman-Menchoff)

Theorem 6 LetU be an open subset ofC andf : U C be a continuous function, f = u+v.

Suppose the partial derivatives of u, v exist and satisfy Cauchy-Riemann equations throughout

U. Then f is holomorphic in U.

Remark 2 We have seen a number of examples of functions that are complex differentiable at

a point but not so at any other points in a neighborhood. As far as the differentiation theory

is concerned such functions are not of much use to us. We would like to concentrate on those

functions which are continuously differentiable in some non empty open subsets. Such functionswill be called holomorphic. Once again, we emphasize the fact that holomorphic functions are

those which have continuous first order partial derivatives satisfying C-R equations in a non

empty open subset ofC. However, in practice, using C-R equations to see whether a function

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is holomorphic or not, would be the last thing that we would like to do. We should have a large

class of holomorphic functions readily known to us and then often a new one could be expressed

in some nice way in terms of these known ones. To this end, the next two lectures will be

devoted to the task of getting familiar with the so called elementary functions.

Exercises:

1. Write down all possible expressions for the Cauchy derivative of a complex function f =

u + v in terms of partial derivatives of u and v.

2. Write down as many details as possible for the proof of if part of theorem 5.

3. Verify that the functions z, z, |z| and z do not satisfy CR equations.

4. Show that the function f(z) = zz is complex differentiable only at z = 0 and find f

(0).

Some optional Material for those of you who can

The operators z and z.Consider the following differential operators:

z

:= 12

x i

y

;

z:= 1

2

x

+ i y

;

We claim that for f = u + iv the following three conditions are equivalent to each

other.

(i)f satisfies CR equations. (ii) fz

= 0. (iii) fz

= 0.

Sol:f

z =1

2(ux + uy + vx vy) = 0 iff CR equations hold. This proves equivalence of(i) and (ii). The proof of equivalence of (i) and (iii) is similar.

Next we establish chain rule in terms of these operators.

Now let f : 1 2, g : 2 1 be any two functions with partial derivatives.

We take the variables in the domain 1 to be z = x + iy and those in 2 to be

w = u + iy. For brevity of exposition let us denote fz

by fz etc. Then the chain rule

can be stated as follows:

(g f)z = gwfz + gwfz; (g f)z = gwfz + gwfz ().

By symmetry, it is enough to establish the first one of (). So, we put f(z) = w =

u + iv. Then by definition, we have

2(g f)z = (g f)x i(g f)y = (guux + gvvx) i(guuy + gvvy).

4gwfz = (gu igv)(fx ify) = gufx gvfy i(gufy + gvfx).

4gwfz = (gu + igv)(fx ify) = gufx + gvfy i(gufy gvfx).

Now adding the last two identities and observing that

fx + fx = 2ux; fy + fy = 2uy; fx fx = 2ivx, fy fy = 2ivy;

we obtain the result.

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Remark 3 The computations that we have carried out has the following interpre-

tation: Let f(T, S) be a polynomial (or a power series) in two variables S, T over

the complex numbers. Let g(z) = f(z, z) be the function obtained by substitution.

Then

gz =f

T (z, z).

We shall illustrate the usefulness of this in the following example.