Chapter 5: DIfferentiation Flashcards
DEF: of a limit
Let f be defined on some punctured neighbourhood of z0 [i.e., it is defined on {z ∈ C : 0 < |z − z_0| < δ} for some δ > 0]. We use the phrase “f(z) → l as z → z_0” to mean that |f(z)−l| → 0 as |z−z_0| → 0.
Ie f(z) tends to l as z tends to z_0 along any path approaching z_0 and the limit l, does not depend on the path chosen.
Since |f(z)- l | and |z-z_0| are real valued :)
Limit is 2 dimensional: limit must exist in all directions towards z_0
To prove limit does not exist, sufficient to find 2 paths
going to z_0 along which the limits exist and are not equal.
Limit of |z|/z as z tends to 0
|z|/z has no limit as z tends to 0
(If exists: limit is 2 dimensional, limit must exist in all directions towards z_0)
|z|/z = 1 for z real and positive
|z|/z = -1 for z real and negative.
Hence there is no l st |z|/z tends to l as z tends to 0
INEQUALITIES
|u − u₀ |
|v − v₀ |
≤ |w − w₀ | = |u + iv − (u₀ + iv₀ )| ≤ |u − u₀ | + |v − v₀
FOR w=u+iv
to tend to
w₀=u₀ +iv₀
as z tends to z₀
(functs of z)
for w=u+iv and w₀=u₀ +iv₀
then w tends to w₀ as z tends to z₀ IFF u tends to u₀ AND v tends to v₀ as z tends to ₀
DEF 5.2
function of z complex tends to value that f(z₀ ) gives as z tends to z₀
If f is defined on a neighbourhood (no longer punctured) of z₀ and
f(z) → f(z₀ ) as z → z₀ , then f is said to be continuous at z₀ .
DEF 5.3 f is differentiable at z₀ if …
the derivative of f at z₀ ,
Let f be a complex valued function of the complex variable z.
Suppose f is defined on a neighbourhood of z₀.
We say that f is differentiable at z₀ if
lim as z→z₀ of
[(f(z) − f(z₀ )) \ (z − z₀ ) ]
exists.
The limit is the derivative of f at z₀ , denoted f’(z₀).
DEF 5.4:
analytic
on a region
The function f is said to be analytic on a region D if f is differentiable
at all points of D.
analytic for regions (by definition, are open sets)
DEF 5.5:
analytic
at z₀
If z₀ is a point, we say that f is analytic at z₀ if f is analytic on some
region containing z₀ .
- so if f is analytic at z₀ then its not only differentiable at z₀ but all points within some open disc to non zero radius centred at z₀ .
- *f is differentiable at z₀ and also at all points sufficiently close to z₀
open sets are natural domains of analytic functions since a function must be defined on a neighbourhood of each point of D
Familiar results from R which are true in C:
about differentiable function
**differentiability implies continuity;
**the sum and product of two differentiable functions is differentiable and the familiar
rules for derivatives of sums and products continue to hold;
**for all non-negative integers n, zⁿ is differentiable and its derivative is nzⁿ⁻¹;
**an analytic function f(w(z)) of an analytic function is analytic on some region and
the chain rule df/dz = (df/dw)(dw/dz) holds.
**if the functions f and g are both differentiable at the point z₀
AND
g(z₀) ≠ 0,
then the quotient f(z)/g(z)
is differentiable at z₀ and its derivative at z₀ is
(f₀’(z₀) g(z₀) − f(z₀) g’(z₀)) /
([g(z₀)]²)
If g(z₀) = 0, then quotient f(z)/g(z) is not defined at z₀ and NOT differentiable at z₀.
USE THESE TO FIND WHERE:
functions involving quotients are analytic, BY:
firstly finding where they are differentiable.
EXAMPLE: Let f(z)= Re z
i.e. f(z) = f(x + iy) = x
where is this funct differentiable?
(continuous
everywhere but nowhere differentiable.)
Then there is no point of C at
which f is differentiable.
**Firstly let z →z₀ along a line parallel to the real axis
write z₀ = x₀ + iy₀ and z =x+iy₀, where x ≠ x₀.
Then
[f(z) - f(z₀) ] / [ z- z₀] = [x-x₀]\ [x-x₀] = 1.
Thus as z →z₀ along a line parallel to the real axis.
[f(z) - f(z₀) ] / [ z- z₀] → 1
** Secondly, let z →z₀ along a line parallel to the IMAGINARY axis
write z₀ = x₀ + iy₀ and z =x₀+iy, where y ≠ y₀.
Then
[f(z) - f(z₀) ] / [ z- z₀] = [x₀-x₀]\ [i(y-y₀)] = 0.
Thus as z →z₀ along a line parallel to the imaginary axis.
[f(z) - f(z₀) ] / [ z- z₀] → 0
** we see that [f(z) - f(z₀) ] / [ z- z₀] doesn’t tend to and limit as z →z₀.
Hence the function f is not differentiable at z₀.
This is true for all points z₀ ∈C.
** Let f(z) = Re z = u +iv and let z=x+iy. then f(z) = f(x+iy) = x and so u(x,y) =x and v(x,y) =0 thus uₓ≠ v_y at all points of C. Hence Rez isnt analytic on C ( and not differentiable at any point of C) . complex function which is continuous at every point of C and is not differentiable at any point.
TMH 5.6 : The Cauchy-Riemann Equations (*)
required for functs to be differentiable ie for appropriate limit to exist
Let f(z) = f(x+iy) = u(x, y)+i v(x, y), where u and v are real valued functions, and let z₀ = x₀ + iy₀. If f is differentiable at z₀, then u and v satisfy the relations
uₓ = v_y, u_y = −vₓ
at (x₀, y₀),(∗)
for PARTIAL DIFFs
(i.e. ∂u/ ∂x = ∂v/ ∂y ,
∂u/∂y = − ∂v/∂x (∗)
at (x₀, y₀).)
In the complex form; “If the function f is differentiable at z₀, then
i∂f/∂x = ∂f/∂y at z₀.
(*) the two forms are equiv as taking i∂f/∂x =∂f/∂y , gives i(uₓ + ivₓ) = u_y + iv_y. Equating real and im gives them
PROOF OF
TMH 5.6 : The Cauchy-Riemann Equations (*)
PROOF
[ f(z) - f(z₀) ]/ [z-z₀]
As f is differentiable at z₀, f must be defined on a neighbourhood of z₀ and compute df/dz at z₀ by letting z → z₀ in any direction.
**firstly let z → z₀ along a line parallel to the real axis.
write z₀ = x₀ + iy₀ and z =x+iy₀, where x ≠ x₀. Then
( f(z) - f(z₀))/ (z-z₀)
= ([u(x,y₀) + iv(x,y₀)] - [u(x₀,y₀) + iv(x₀,y₀)] ) / x-x₀
= [u(x,y₀) - u(x₀,y₀) ] / x-x₀
+
i [v(x,y₀)- v(x₀,y₀)]/[x-x₀]
→uₓ(x₀,y₀) + ivₓ(x₀,y₀)
as x→x₀
Thus
f’(z₀) = limit as z →z₀
uₓ(x₀,y₀) + ivₓ(x₀,y₀)
**next let z → z₀ along a line parallel to the imaginary axis.
write z₀ = x₀ + iy₀ and z =x₀+iy, where y ≠ y₀.
( f(z) - f(z₀))/ (z-z₀)
= ([u(x₀,y) + iv(x₀,y)] - [u(x₀,y₀) + iv(x₀,y₀)] ) / i(y-y₀)
= [u(x₀,y) - u(x₀,y₀) ] / i(y-y₀)
+
[v(x₀,y)- v(x₀,y₀)]/[ (y-y₀)]
→-iu_y(x₀,y₀) + v_y(x₀,y₀)
as y→y₀
Thus
f’(z₀) = limit as z →z₀
iu_y(x₀,y₀) + v_y(x₀,y₀)
EQUATING REAL AND IMAGINARY PARTS
gives
uₓ(x₀, y₀) = v_y(x₀, y₀), vₓ(x₀, y₀) = −u_y(x₀, y₀)
ie
∂u/∂x = ∂v/∂y ,
∂v/∂x = −∂u/∂y at (x₀,y₀)
ie
f’(z₀)= [uₓ + ivₓ]ₓ₌ₓ₀,y₌y₀ = [∂f/∂x ]ₓ₌ₓ₀,y₌y₀
f'(z₀) = [-iu_y + v)y]ₓ₌ₓ₀,y₌y₀ -i = [∂f/∂y ]ₓ₌ₓ₀,y₌y₀
HENCE at z₀ = x₀ + iy₀,
∂f/∂x = −i∂f/∂y .
PROPERTIES of The Cauchy-Riemann Equations (*)
** four versions of f'. if f= u +iv is analytic f'(z) = uₓ + ivₓ = uₓ - iu_y =v_y - iu_y =u_y + ivₓ
** the cauchy- riemann eq are AXIS DOMINATED- only make statements about the appropriate limit in 2 directions, along lines parallel to real and imaginary axes.
(lim as z→z₀ of
[(f(z) − f(z₀ )) \ (z − z₀ ) ] )
We know derivative exists at z₀ if this limit exists for all approaches to it. Hence dont expect the converse to be true.
(there are examples in which the Cauchy-Riemann equations hold at a point
z₀ but the function is not differentiable there.)
the fact that the Cauchy-Riemann equations are satisfied at a certain
point is NOT SUFFICIENT to guarantee differentiability at that point- must be continuous before checking C-R
EXAMPLE:
z₀ = 0 and let f be defined by
f(z) =
{1 on the axes
{ 0 elsewhere
is it continuous/differentiable at the originwrite f(z) = u(x,y) + iv(x,y) where u,v are real valued so that v(x,y) = 0, u(x,0)= 1 u(0,y) = 1 for all real x,y. Then
uₓ(0,0) = lim x→x₀ of
[[u(x,0) - u(0,0)]/x] = 0
similarly
u_y(0,0) = 0
vₓ (0,0) = 0
v_y (0,0) = 0
and the C-R equations are satisfied at the origin but f isnt continuous there so not differentiable
EXAMPLE:
Prove that, f analytic on C and Re f constant on C implies that f is constant on
C.
f(z) = f(x + iy) = u(x, y) + iv(x, y), where u, v are
real-valued. here u = Re f is constant — uₓ = u_y = 0 everywhere.
in aaddition function f is analytic in C and so C-R equ
uₓ = v_y
vₓ = − u_y are satisfied everywhere.
So v is independent of x and y. Thus v is constant, so f = u+iv is constant.
