Limits and Continuity
JunFeng Yin
Definition 1
Let f be a function defined in a neighborhood of z0 . Then f is
continuous at z0 if
lim f (z) = f (z0 ).
z→z0
A function f is said to be continuous on a set S if it is continuous
at each point of S.
Theorem 2
If limz→z0 f (z) = A and limz→z0 g (z) = B, then
1. limz→z0 (f (z) ± g (z)) = A ± B
2. limz→z0 f (z)g (z) = AB
3. limz→z0
f (z)
g (z)
=
A
B
ifB 6= 0
Theorem 3
If f (z) and g (z) are continuous at z0 , then so are f (z) ± g (z) and
f (z)g (z). The quotient f (z)/g (z) is also continuous at z0
provided g (z0 ) 6= 0.
Polynomial function:
a0 + a1 z + a2 z 2 + · · · + an z n ,
are continuous on the whole plane C.
Rational function:
a0 + a1 z + a2 z 2 + · · · + an z n
b0 + b1 z + b2 z 2 + · · · + bm z m
are continuous at each point where the denominator does not
vanish.
Example 4
Find the limit, as z → 2i, of the function f1 (z) = z 2 − 2z + 1,
f2 (z) = (z + 2i)/z and f3 (z) = (z 2 + 4)/z(z − 2i).
(-3-4i,2)
Remark A function has a removable discontinuity at z0 if it can
be defined or redefined at a single point z0 so as to be continuous
at z0 .
e.g., define f3 (2i) = 2.
Remark:
I
limz→z0 f (z) = ∞ means ∀M > 0, ∃δ > 0 s.t. |f (z)| > M
whenever 0 < |z − z0 | < δ.
I
A complex number approach infinity when their magnitudes
approach infinity.
Example:
lim
z→3i
z
= ∞,
z2 + 9
iz − 2
i
= ,
z→∞ 4z + i
4
lim
z 3 + 3i
= ∞.
z→∞ z 2 + 5i
lim
§3 Analyticity
We are ready to turn to the main topic of this book
the theory of analytic function.
z = x + yi → f (z) = u(x, y ) + v (x, y )i
What the difference between
u(x, y ) = x 2 − y 2 ,
v (x, y ) = 2xy ,
u(x, y ) = x 2 − y 2 ,
v (x, y ) = 3xy ,
and
z = x + yi → f (z) = u(x, y ) + v (x, y )i
What the difference between
u(x, y ) = x 2 − y 2 ,
v (x, y ) = 2xy ,
u(x, y ) = x 2 − y 2 ,
v (x, y ) = 3xy ,
and
It is seen that
f (z) = u(x, y ) + v (x, y )i = x 2 − y 2 + 2xyi = (x + yi)2
However, the second one can not be written in terms of z.
Admissible
1. z = x + yi
2. z 2 = x 2 − y 2 + 2xyi
3. z 3 = x 3 − 3xy 2 + (3x 2 y − y 3 )i
4. 1/z =
x
x 2 +y 2
y
− i x 2 +y
2
5. e z = 1 + z + z 2 /2! + z 3 /3! + · · ·
Inadmissible
1. Re(z) = x
2. Im(z) = y
3. x 2 − y 2 + 3xyi
4. z = x − yi
p
5. |z| = x 2 + y 2
Since
z +z
z −z
, x=
2
2
any function can be expressed in terms of z and z.
x=
Since
z +z
z −z
, x=
2
2
any function can be expressed in terms of z and z.
For instance,
x=
x 2 − y 2 + 3xyi
=
=
(z + z)2 (z − z)2
z +z z −z
−
+3
i
4
4
2
2
5 2 1 2
z − z
4
4
Example 5
Express the following functions in terms of z and z.
f1 (z) =
x − 1 + iy
(x − 1)2 + y 2
f2 (z) = x 2 + y 2 + 3x + 1 + 3yi
Example 5
Express the following functions in terms of z and z.
f1 (z) =
x − 1 + iy
(x − 1)2 + y 2
f1 (z) =
=
=
f2 (z) = x 2 + y 2 + 3x + 1 + 3yi
x − 1 + iy
(x − 1)2 + y 2
z+z
z−z
2 −1− 2 i
z−z 2
2
( z+z
2 − 1) + ( 2 )
z −1
1
=
zz − z − z + 1
z −1
(z + z)2 (z − z)2
z +z
z −z
+
+ 3(
) + 1 + 3(
)i
4
4i 2
2
2
= zz + 3z + 1
f2 (z) =
Definition 6
Let f be a complex-valued function defined in a neighborhood of
z0 . Then the derivative of f at z0 is given by
f (z0 + ∆z) − f (z0 )
df
(z0 ) ≡ f 0 (z0 ) := lim
∆z→0
dz
∆z
provided this limit exists. (Such an f is said to be differentiable
at z0 .)
Example 7
Express f (z) = z is nowhere differentiable.
Example 7
Express f (z) = z is nowhere differentiable.
Solution The difference quotient for this function takes the form
f (z0 + ∆z) − f (z0 )
z0 + ∆z − z0
∆z
=
=
∆z
∆z
∆z
Now if ∆z → 0 through real value, then ∆z = ∆x and ∆z = ∆x.
so the difference quotient is 1.
On the other hand, if ∆z → 0 through y-axis, then ∆z = ∆yi and
∆z = −∆yi. so the difference quotient is −1.
Hence, z is not differentiable.
Neither x, y , nor |z| is differentiable.
Example 8
Show that, for any positive integer n,
d n
z = nz n−1 .
dz
Example 8
Show that, for any positive integer n,
d n
z = nz n−1 .
dz
Solution Using the binomial formula, we find
nz n−1 ∆z + 12 n(n − 1)z n−2 (∆z)2 + · · · + (∆z)n
(z + ∆z)n − z n
=
∆z
∆z
Thus
d n
1
z = lim [nz n−1 + n(n−1)z n−2 (∆z)+· · ·+(∆z)n−1 ] = nz n−1 .
∆z→0
dz
2
It is the same as for the real-variable case.
Theorem 9
If f and g are differentiable at z, then
(f + g )0 (z) = f 0 (z) + g 0 (z),
(cf )0 (z) = cf 0 (z),
(∀c)
(fg )0 (z) = f (z)g 0 (z) + f 0 (z)g (z),
f
g (z)f 0 (z) − f (z)g 0 (z)
,
( )0 (z) =
g
g 2 (z)
if g (z) 6= 0
If g is differentiable at z and f is differentiable at g (z), then the
chain rule holds:
d
f (g (z)) = f 0 (g (z))g 0 (z)
dz
Example 10
Compute the derivative of
f (z) =
z2 − 1
z2 + 1
100
.
Example 10
Compute the derivative of
f (z) =
z2 − 1
z2 + 1
100
.
Solution Unless z = ±i (where the denominator is zero), the usual
calculus rules apply. Thus
0
f (z) = 100
z2 − 1
z2 + 1
99
(z 2 + 1)2z − (z 2 − 1)2z
(z 2 − 1)99
=
400z
(z 2 + 1)2
(z 2 + 1)101
Definition 11
A complex-valued function f (z) is said to be analytic on an open
set G if it has a derivative at every point of G .
I
f (z) is said to be analytic at the point z0 means that f (z) is
analytic in some neighborhood of z0 .
I
A point where f is not analytic but which is the limit of points
where f is analytic is known as a singular point or
singularity.
I
A rational function of z is analytic at every point for which its
denominator is nonzero.
I
If f (z) is analytic on the whole complex plane, it is said to be
entire. e.g, all polynomial functions of z are entire.
§4 The Cauchy-Riemann Equations
Denote f (z) = u(x, y ) + v (x, y )i is differentiable at z0 = x0 + y0 i,
then from
f (z0 + ∆z) − f (z0 )
f 0 (z0 ) = lim
,
∆z→0
∆z
we have
∂u
∂v
f 0 (z0 ) =
(x0 , y0 ) + i (x0 , y0 ),
∂x
∂x
and
∂u
∂v
f 0 (z0 ) = − i(x0 , y0 ) +
(x0 , y0 ),
∂y
∂y
thus
∂u
∂v
=
,
∂x
∂y
∂u
∂v
=−
∂y
∂x
Theorem 12
A necessary condition for a function f (z) = u(x, y ) + v (x, y )i to
be differentiable at a point z0 is that the Cauchy-Riemann
equations hold at z0 .
Consequently, if f is analytic in an open set G , then the
Cauchy-Riemann equations must hold at every point of G .
Example 13
Show f (z) = (x 2 + y ) + (y 2 − x)i is not analytic at any point.
Solution Since u(x, y ) = x 2 + y and v (x, y ) = y 2 − x, we have
∂u
∂x
∂u
∂y
= 2x,
= 1,
∂v
= 2y
∂y
∂v
= −1
∂x
Hence, the C-R equation are simultaneously satisfied only on the
line x = y , and therefore in no open disk.
Thus, f (z) is nowhere analytic.
Theorem 14
Let f (z) = u(x, y ) + v (x, y )i be defined in some open set G
containing the point z0 . If the first partial derivatives of u and v
exist in G , are continuous at z0 , and satisfy the C-R equations at
z0 , then f is differentiable at z0 .
Consequently, if the first partial derivative are continuous and
satisfy the C-R equations at all points of G , then f is analytic in G .
Example 15
Prove that the function f (z) = e z = e x cos y + ie x sin y is entire,
and find its derivative.
Solution It is easy to verify the first partial derivatives are
continuous and satisfy C-R equation.
f 0 (z) =
∂u ∂u
+
i = e x cos y + e x sin yi
∂x
∂y
Theorem 16
If f (z) is analytic in a domain D and if f 0 (z) = 0 everywhere in D,
then f (z) is constant in D.
Denote u and v in terms of polar coordinates (r , θ), show that the
C-R equations can be written in the form
1 ∂v
∂u
=
,
∂r
r ∂θ
∂v
1 ∂u
=−
,
∂r
r ∂θ
§5 Harmonic Function
Consider the Laplace equation
∇2 φ :=
∂2φ ∂2φ
+
=0
∂x 2
∂y 2
Definition 17
A real-valued function φ(x, y ) is said to be harmonic in a domain
D if all its second-order partial derivatives are continuous in D and
if, at each point of D, φ satisfies Laplace’s equation.
Theorem 18
If f (z) = u(x, y ) + iv (x, y ) is analytic in a domain D, then each of
the functions u(x, y ) and v (x, y ) is harmonic in D.