By the Mean-Value Theorem (for real functions of a real variable) u(z + h

Analyticity and the Cauchy Riemann Equations

Proposition Suppose f_x and f_y exist in a neighborhood of z. If f_x and f_y are continuous at z and f_y = if_x there, then f is differentiable at z.

Proof Let f = u + iv, h = ξ + iη. By the Mean-Value Theorem (for real functions of a real variable)

u(z + h) − u(z)

h = u(x + ξ, y + η) − u(x, y) ξ + iη

= u(x + ξ, y + η) − u(x + ξ, y)

ξ + iη +u(x + ξ, y) − u(x, y) ξ + iη

= η

ξ + iηu_y(x + ξ, y + θ₁η) + ξ

ξ + iηu_x(x + θ₂ξ, y)

= η

ξ + iηu_y(z₁) + ξ

ξ + iηu_x(z₂) and

v(z + h) − v(z)

h = η

ξ + iηv_y(x + ξ, y + θ₃η) + ξ

ξ + iηv_x(x + θ₄ξ, y)

= η

ξ + iηv_y(z₃) + ξ

ξ + iηv_x(z₄) for some 0 < θ_k < 1, k = 1, 2, 3, 4. Thus

f (z + h) − f (z)

h = η

ξ + iηu_y(z₁) + iv_y(z₃) + ξ

ξ + iηu_x(z₂) + iv_x(z₄) Since f_y = if_x at z,

f_x(z) = ξ + iη

ξ + iηf_x(z) = ξ

ξ + iηf_x(z) + η

ξ + iηf_y(z), and thus

f (z + h) − f (z)

h − f_x(z) = f (z + h) − f (z)

h − ξ

ξ + iηf_x(z) − η

ξ + iηf_y(z)

= η

ξ + iηu_y(z₁) − u_y(z) + i v_y(z₃) − v_y(z)


+ ξ

ξ + iηu_x(z₂) − u_x(z) + i v_x(z₄) − v_x(z)
.

Since lim

h→0|z_k − z| = 0 for each 1 ≤ k ≤ 4,    

η ξ + iη

    ,

η ξ + iη

≤ 1 and f_x = u_x + iv_x and f_y = u_y+ iv_y are continuous at z, we have

h→0lim

f (z + h) − f (z)

h − f_x(z) = 0.

Examples

(a) Let f (z) = |z|² = x²+ y², z ∈ C.

Since fx = 2x, fy = 2y are continuous for all z ∈ C, f is differentiable if and only if f^y = ifx

by the above proposition. Hence f is differentiable only at z = 0.

(b) The converse of the above proposition is not true. For example, consider

f (x, y) =







xy(x + iy)

x²+ y² if z 6= 0

0 if z = 0.

Note that f = 0 on both axes so thatf_x = f_y = 0 at the origin but since limz→0

f (z) − f (0)

z = lim

(x,y)→(0,0)

xy x²+ y²

y=αx= α

1 + α² depends on α, the limit does not exist and f is not differentiable at z = 0. In fact,f_x and f_y

f_x(x, y) =







2xy³+ iy²(y²− x²)

(x²+ y²)² if z 6= 0

0 if z = 0,

f_y(x, y) =







x²(x²− y²) + 2ix³y

(x² + y²)² if z 6= 0

0 if z = 0

are not continuous at z = 0.

(c) Consider the real-valued function f : R² → R defined by

f (x, y) =







x²+ y²
sin

 1

x²+ y²



if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0).

Note that f is real-differentiable at every point (x, y) ∈ R²,

f_x(x, y) =





 2x sin

 1

x² + y²



− 2x

x²+ y² cos

 1

x² + y²



if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0),

and

f_y(x, y) =





 2y sin

 1

x²+ y²



− 2y

x²+ y² cos

 1

x²+ y²



if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

with f_x(0, 0) = 0 = f_y(0, 0), but f_x, f_y are not bounded (hence not continuous) in a neighborhood of (0, 0).

Remark (An algebraic interpretation of the Cauchy-Riemann equations) Consider the complex polynomial

p(z) =

n

X

k=0

c_kz^k.

Look at complex-valued polynomials of real variables x, y:

p(x, y) =

n

X

m=0

X

k+`=m,k, `≥0

c<sub>k, `</sub>x<sup>k</sup>y<sup>`</sup> with c<sub>k, `</sub>= ∂<sup>k+`</sup>p

∂x^k∂y^`(0, 0) (∗)

and take a ”new basis” as z = x + iy and ¯z = x − iy. Write the polynomial in z, ¯z:

p(z, ¯z) =

n

X

m=0

X

k+`=m,k, `≥0

d<sub>k, `</sub>z<sup>k</sup>z¯<sup>`</sup> (†)

We claim that there’s a 1 − 1 correspondence between polynomials given by (∗) and (†). In fact, since z = x + iy and ¯z = x − iy implies that x = z + ¯z

2 , y = z − ¯z

2i and by the Chain Rule

∂

∂z = ∂x

∂z

∂

∂x +∂y

∂z

∂

∂z = 1 2

∂

∂x − i ∂

∂y
, ∂

∂ ¯z = ∂x

∂ ¯z

∂

∂x +∂y

∂ ¯z

∂

∂y = 1 2

∂

∂x + i ∂

∂y
.

Claim We have the equalities

∂

∂ ¯z(z^kz¯<sup>`</sup>) = `z^kz¯^`−1

∂

∂z(z^kz¯^{`</sup>) = kz<sup>k−1</sup>z¯<sup>`} Proof of Claim We simply check

∂

∂ ¯z(z^kz¯^`) = 1 2

∂

∂x + i ∂

∂y
(x + iy)^k(x − iy)^`

= 1

2k(x + iy)^k−1(x − iy)<sup>`</sup>+ `(x + iy)^k(x − iy)^`−1 + i

2ki(x + iy)^k−1(x − iy)<sup>`</sup>− i`(x + iy)^k(x − iy)^`−1

= 1

2kz^k−1z¯<sup>`</sup>+ `z^kz¯^`−1 + i

2kiz^k−1z¯<sup>`</sup>− i`z^kz¯^`−1

= `z<sup>k</sup>z¯<sup>`−1</sup>

Claim (equivalent form of CR equations) f (z) satisfies CR equations ⇐⇒ ∂

∂ ¯zf = 0.

Proof of Claim Since

∂f

∂ ¯z = 1 2

∂f

∂x + i∂f

∂y

=⇒ f_y = if_x ⇐⇒ ∂f

∂ ¯z = 0.

Remark The conclusion is that

p(z, ¯z) =

n

X

m=0

X

k+`=m, k, `≥0

d<sub>k, `</sub>z<sup>k</sup>z¯<sup>`</sup>

is holomorphic if and only if d<sub>k, `</sub>= 0 for ` > 0 (e.g. no ¯z’s). Alternatively, p(x, y) =

n

X

m=0

X

k+`=m, k, `≥0

c<sub>k, `</sub>x<sup>k</sup>y<sup>`</sup>

is holomorphic if it can be written as p(z) =

n

X

k=0

c_kz^k.

Let’s give some basic hints that holomorphic functions behave a bit like power series:

Proposition

(a) If f = u + iv is analytic in a region D and u is constant, then f is constant.

Proof Since u is constant,

u_x = u_y = 0 in D, and, by the Cauchy-Riemann equations,

v_x = v_y = 0 in D.

By the Mean Value Theorem, u and v are constants along any line segment in D. So f = u + iv is constant on D.

(b) If f is analytic in a region D and if |f | is constant there,then f is constant.

Proof If |f | = 0, the proof is immediate. Otherwise u²+ v² ≡ C 6= 0 on D.

Taking the partial derivatives with respect to x and y, we see that u u_x+ v v_x ≡ 0

u u_y+ v v_y ≡ 0 Making use of the Cauchy-Riemann equations, we obtain

u ux− v uy ≡ 0 v u_x+ u u_y ≡ 0 so that

(u²+ v²)u_x ≡ 0 on D

and u_x = v_y ≡ 0. Similarly, u_y and v_x are identically zero, hence f is constant.

The Functions e^z, sin z, cos z, log z, z^α

In general, either one of the following methods will be used to define a complex analogue of a real analytic function.

(i) we determine the same fundamental defining properties;

(ii) we use its power series to obtain a holomorphic extension.

Definitions

(1) To define an extension f (z) = e^z of the real exponential function f (x) = e^x, we observe that

(i) e⁰ = 1, d

dxe^x = e^x, . . . , which implies for each fixed x₁ ∈ R, d

dx e^x+x1

e^x = e^xe^x+x1 − e^x+x1e^x

(e^x)² = 0 =⇒ e^x+x1

e^x = constant = e^x1 =⇒ e^x+x1 = e^xe^x1 implies that f (x) = e^x is infinitely differentiableand satisfies that

f (0) = 1, df

dx = f and f (x + x₁) = f (x)f (x₁) for all x, x₁ ∈ R.

(ii) e^x=

∞

X

k=0

x^k

k! is convergent for all x ∈ R,implies that f (x) = e^x is a real analytic function defined by a convergent power series for all x ∈ R.

So we should define f (z) such that

(i) f (x) = e^x for all z = x + i 0 ∈ R and f (z1+ z₂) = f (z₁)f (z₂) for all z₁, z₂ ∈ C.

This implies that

f (x + iy) = f (x)f (iy) = e^xf (iy) ∀x, y ∈ R and, by setting x = 0 and using f (0) = 1, we have

f (iy) = A(y) + iB(y),

where A, B are real-valued functions satisfying A(0) = 1, B(0) = 0, and f (x + iy) = e^xA(y) + ie^xB(y).

Using the Cauchy-Riemann equations f_y = if_x, A(0) = 1 and B(0) = 0, we get (A⁰(y) = −B(y)

B⁰(y) = A(y) =⇒

(A⁰⁰(y) + A(y) = 0

B⁰⁰(y) + B(y) = 0 =⇒ A(y) = cos y and B(y) = sin y.

This implies that f (z) = e^z should be defined by

e^z = e^x(cos y + i sin y) ⇐⇒ e^iy = cos y + i sin y for all z = x + iy ∈ C.

(ii) f (z) =

∞

X

k=0

z^k

k! for all z ∈ C.

Since

lim

k→∞

1 k!

1/k

= 0,

the radius of convergence is ∞, f (z) =

∞

X

k=0

z^k

k! is convergent and differentiable for all z ∈ C. Since

f (0) = 1, f⁰(z) = f (z), · · · ∀ z ∈ C, by the term-by-term differentiation theorem, and

d dz

f (z + w)

f (z) = f (z)f (z + w) − f (z + w)f (z)

[f (z)]² = 0 for a fixed w ∈ C,

we obtain that

f (z)f (w) = f (z + w) for all z, w ∈ C which implies that f (z) =

∞

X

k=0

z^k

k! = e^z for all z ∈ C.

(2) Let f (z) = sin z, cos z for z ∈ C.

Since

e^iz = 1 + (iz) + (iz)²

2! + · · · = 1 − z² 2! +z⁴

4! + · · ·
+ i z − z³ 3! +z⁵

5! + · · ·
, we have

e^ix =

∞

X

k=0

(−1)^k

(2k)! x^2k + i

∞

X

k=0

(−1)^k

(2k + 1)!x^2k+1=: cos x + i sin x for x ∈ R and

cos x = e^ix+ e^−ix

2 and sin x = e^ix − e^−ix

2i for x ∈ R.

So sin z, cos z should be defined by cos z = e^iz+ e^−iz

2 and sin z = e^iz− e^−iz

2i for z ∈ C.

We can verify the usual properties that

∂

∂z cos z = − sin z, ∂

∂z sin z = cos z,

cos 0 = 1, sin 0 = 0, cos(−z) = cos z, sin(−z) = − sin z, etc.

Check the addition formulas for cos(z₁ + z₂) and sin(z₁+ z₂).

(3) Let w = log z ⇐⇒ z = e^w for z ∈ C.

For w = a + bi ∈ C, since

z = e^w = e^a(cos b + i sin b) = re^iθ,

log z should be defined by

log z = log r + iθ + i(2πn), z ∈ C so the complex log is multi-valued.

By writing z = re^iθ = r(cos θ + i sin θ), we define the principal branch of log z as follows.

For each z ∈ C \ {z = x + 0 · i ∈ C | x ≤ 0}, the principal branch of log z is defined by log z = log |z| + iArg(z), where Arg(z) ∈ (−π, π).

So the principal branch of log z has imaginary part in (−π, π).

Claim log z is holomorphic given a branch cut, and

∂

∂z log z
= 1

z for all z ∈ C \ {z = x + 0 · i ∈ C | x ≤ 0}.

Proof of Claim Cosider

h→0lim

log(z + h) − log z

h .

Setting w1 = log(z + h), w = log z, since log is continuous given a branch cut, we have lim

h→0w₁ = w and thus

lim

h→0

log(z + h) − log z

h = lim

h→0

w₁− w

e^w1 − e^w = 1

wlim1→w

e^w1 − e^w w₁− w

= 1 e^w = 1

z.

Intuitively, log z looks like a helix above C \ {0}; we call this the Riemann surface for log z.

(4) For α ∈ R, let f (z) = z^α for z ∈ C.

Define

z^α = e^{α log z} = eα(log z+2πni), for z ∈ C

which is potentially multi-valued. e^{α log z} is the principal branch; now examine e^2πnαi. If α /∈ Q, e^2πnαi takes infinitely many values.

If α = k

l ∈ Q, e^2πnαi takes l values (e.g. e^πni= ±1 for α = 1 2).

For example, when considering√

z, we still need a branch cut

√

re^iθ =√

re^iθ/2, θ ∈ (−π, π).