Instructor:

Complex Variables & Transforms (MATH- 232)

Dr. Naila Amir
 Complex Functions and Mappings
Book: A First Course in Complex Analysis with Applications by
Dennis G. Zill and Patrick D. Shanahan.

• Chapter: 2
• Sections: 2.6
 Limits and Continuity
▪ The most important concept in calculus is that of the limit. Recall that
lim 𝑓 𝑥 = 𝐿,
𝑥→𝑥0
intuitively means that values 𝑓(𝑥) of the function f can be made arbitrarily
close to the real number 𝐿 if values of 𝑥 are chosen sufficiently close to, but
not equal to, the real number 𝑥0 .

▪ Complex limits play an equally important role in study of complex analysis. The
concept of a complex limit is similar to that of a real limit in the sense that:
lim 𝑓 𝑧 = 𝐿,
𝑧→𝑧0
means that the values 𝑓(𝑧) of the complex function 𝑓 can be made arbitrarily
close the complex number 𝐿 if values of 𝑧 are chosen sufficiently close to, but
not equal to, the complex number 𝑧0 .
 Real Multivariable Limits
We now study a practical method for computing complex limits in the theorem
presented below. This theorem establishes an important connection between the
complex limit of 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) and the real limits of the real-valued
functions of two real variables 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦). Since every complex function is
completely determined by the real functions 𝑢 and 𝑣, it should not be surprising that
the limit of a complex function can be expressed in terms of the real limits of 𝑢 and 𝑣.

Theorem: Real and Imaginary Parts of a Limit

Suppose that 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦 , 𝑧0 = 𝑥0 + 𝑖𝑦0 , and 𝐿 = 𝑢0 + 𝑖𝑣0 . Then
lim 𝑓(𝑧) = 𝐿
𝑧⟶𝑧0
if and only if
lim 𝑢(𝑥, 𝑦) = 𝑢0 and lim 𝑣(𝑥, 𝑦) = 𝑣0 .
𝑥,𝑦 ⟶(𝑥0 ,𝑦0 ) 𝑥,𝑦 ⟶(𝑥0 ,𝑦0 )
Properties of Complex Limits
 Infinite limit and Limit at infinity
Analogous to real analysis, we can also define the concepts of infinite limits and limits
at infinity for complex functions. Intuitively, the limit:
lim 𝑓 𝑧 = 𝐿 ,
𝑧⟶∞
means that values 𝑓(𝑧) of the function 𝑓 can be made arbitrarily close to 𝐿 if values
of 𝑧 are chosen so that |𝑧| is sufficiently large. A precise statement of a limit at
infinity is:
The limit of 𝑓(𝑧) as 𝑧 tends to ∞ exists and is equal to 𝐿 if for every
𝜀 > 0 there exists a 𝛿 > 0 such that |𝑓(𝑧) − 𝐿| < 𝜀 whenever |𝑧| > 1/𝛿.

Using this definition, it is not hard to show that:

1
lim 𝑓 𝑧 = 𝐿 iff lim 𝑓 = 𝐿.
𝑧⟶∞ 𝑧⟶0 𝑧
 Infinite limit and Limit at infinity
Moreover, we have:
1
lim 𝑓 𝑧 = ∞ iff lim = 𝐿.
𝑧⟶∞ 𝑧⟶0 𝑓(1/𝑧)

Similarly, the infinite limit:

lim 𝑓 𝑧 = ∞ ,
𝑧⟶𝑧0
is defined by:

The limit of 𝑓(𝑧) as 𝑧 tends to 𝑧0 is ∞ if for every 𝜀 > 0 there

is a 𝛿 > 0 such that |𝑓(𝑧)| > 1/𝜀 whenever 0 < |𝑧 − 𝑧0 | < 𝛿.

From this definition we obtain the following result:

1
lim 𝑓 𝑧 = ∞ iff lim = 0.
𝑧⟶𝑧0 𝑧⟶𝑧0 𝑓(𝑧)
 Practice Question
1. Compute:
Im 𝑧 2
lim .
𝑧⟶3𝑖 𝑧 + Re (𝑧)

2. Determine:
3 + 𝑖 𝑧 4 − 𝑧 2 + 2𝑧
lim .
𝑧→𝑖 𝑧+1

3. Determine:
𝑧 2 − 2𝑧 + 4
lim .
𝑧→1+𝑖 3 𝑧−1−𝑖 3
4. Determine:
𝑧 2 + 𝑖𝑧 − 2
lim 2
.
𝑧→∞ (1 + 2𝑖)𝑧
 Continuity of Complex Functions
The definition of continuity for a complex function is, in essence, the same as that for a
real function. That is, a complex function 𝑓 is continuous at a point 𝑧0 if the limit of 𝑓
as 𝑧 approaches 𝑧0 exists and is the same as the value of 𝑓 at 𝑧0 , i.e.,
lim 𝑓 𝑧 = 𝑓 𝑧0 .
𝑧→𝑧0

Criteria for Continuity at a Point:

A complex function 𝑓 is continuous at a point 𝑧0 if each of the following three
conditions hold:
i. lim 𝑓 𝑧 exists,
𝑧→𝑧0

ii. 𝑓 is defined at 𝑧0 , and

iii. lim 𝑓 𝑧 = 𝑓 𝑧0 .
𝑧→𝑧0
 Continuity of Complex Functions
Example: Checking Continuity at a point
Consider the function 𝑓 𝑧 = 𝑧 2 − 𝑖𝑧 + 2. In order to determine if 𝑓 is continuous at

say the point 𝑧0 = 1 − 𝑖, we must find lim 𝑓 𝑧 and 𝑓 𝑧0 , the check to see whether
𝑧→𝑧0

these two complex values are equal. Thus,

lim 𝑓 𝑧 = lim 𝑧 2 − 𝑖𝑧 + 2 = 1 − 3𝑖.

𝑧→𝑧0 𝑧→1−𝑖

Furthermore, for 𝑧0 = 1 − 𝑖 we have:

𝑓 𝑧0 = 𝑓 1 − 𝑖 = 1 − 3𝑖.

Since lim 𝑓 𝑧 = 𝑓 𝑧0 , we conclude that 𝑓 𝑧 is continuous at the point 𝑧0 = 1 − 𝑖.

𝑧→𝑧0
 Real and Imaginary Parts of a Continuous Function
Suppose that 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) and 𝑧0 = 𝑥0 + 𝑖𝑦0 . Then the complex
function 𝑓 is continuous at the point 𝑧0 if and only if both real functions 𝑢(𝑥, 𝑦) and
𝑣(𝑥, 𝑦) are continuous at the point (𝑥0 , 𝑦0 ).

Properties of Continuous Functions

 Continuity of Polynomial & Rational Functions
▪ Polynomial functions are continuous on the entire complex plane ℂ.

𝑝 𝑧
▪ A rational function 𝑓 𝑧 = is quotient of the polynomial functions
𝑞 𝑧
𝑝(𝑧) and 𝑞 𝑧 .

▪ Thus, it follows from above fact about polynomial functions and

properties of continuous functions that 𝑓(𝑧) is continuous at every point
𝑧0 for which 𝑞(𝑧0 ) ≠ 0.

▪ In other words, we say that rational functions are continuous on their

domains.
 Example
Show that the function 𝑓(𝑧) = 𝑧ҧ is continuous on ℂ.
Solution:
According to theorem, 𝑓(𝑧) = 𝑧ҧ = 𝑥 − 𝑖𝑦 is continuous at 𝑧0 = 𝑥0 + 𝑖𝑦0 if
both 𝑢(𝑥, 𝑦) = 𝑥 and 𝑣(𝑥, 𝑦) = −𝑦 are continuous at (𝑥0 , 𝑦0 ). Since,
𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are two-variable polynomial functions, it follows that:

lim 𝑢 𝑥, 𝑦 = 𝑥0 and lim 𝑣 𝑥, 𝑦 = −𝑦0 .

(𝑥,𝑦)→(𝑥0 ,𝑦0 ) (𝑥,𝑦)→(𝑥0 ,𝑦0 )

This implies that 𝑢 and 𝑣 are continuous at (𝑥0 , 𝑦0 ), and 𝑓(𝑧) = 𝑧ҧ is

continuous at 𝑧0 = 𝑥0 + 𝑖𝑦0 . Since 𝑧0 = 𝑥0 + 𝑖𝑦0 is an arbitrary point, so we
conclude that the function 𝑓(𝑧) = 𝑧ҧ is continuous on the entire complex
plane.
 Example
Show that the principal square root function 𝑓(𝑧) = 𝑧1/2 is discontinuous at the point
𝑧0 = −1.
Solution:
We show that 𝑓(𝑧) = 𝑧1/2 is discontinuous at 𝑧0 = −1 by demonstrating that
lim 𝑓(𝑧) = lim 𝑧1/2
𝑧⟶𝑧0 𝑧⟶−1
does not exist. In order to do so, we present two ways of letting 𝑧 approach −1 that yield
different values of this limit. The principal square root function is defined by:
θ
𝑖2
𝑧1/2 = 𝑟𝑒 ,
where 𝑟 = |𝑧| and 𝜃 = Arg 𝑧 . Now consider 𝑧 approaching −1 along
the quarter of the unit circle lying in the second quadrant, i.e., consider
the points 𝑟 = 1, 𝜋/2 < 𝜃 < 𝜋. Then we have:
θ θ
𝑖2 𝑖2
lim 𝑧1/2 = lim 1𝑒 = lim 𝑒 = 𝑖. (1)
𝑧⟶−1 𝑧⟶−1 θ⟶𝜋
Solution:
Next, we let 𝑧 approach −1 along the quarter of the unit circle lying in
the third quadrant. Along this curve we have the points 𝑧 = 𝑒 𝑖𝜃 ,
−𝜋 < 𝜃 < −𝜋/2, with 𝜃 approaching −𝜋. In this case we find:
θ θ
𝑖2 𝑖2
lim 𝑧1/2 = lim 1𝑒 = lim 𝑒 = −𝑖 . (2)
𝑧⟶−1 𝑧⟶−1 θ⟶−𝜋

From (1) and (2) we observe that lim 𝑧1/2 does not exist.
𝑧⟶−1
Therefore, we conclude that the principal square root function 𝑓(𝑧) = 𝑧1/2 is discontinuous
at the point 𝑧0 = −1.
Note: By using the same procedure, we can easily show that the principal square root
function is discontinuous for all points on the negative real axis. If we exclude the negative
real axis from the domain of the principal square root function i.e., if we have −𝜋 < 𝜃 < 𝜋
then 𝑓 𝑧 = 𝑧1/2 , −𝜋 < 𝜃 < 𝜋 is continuous for all values of 𝑧 in the indicated domain.
 Branches
A multiple-valued function 𝐹(𝑧) assigns a set of complex numbers to the input 𝑧. In
practice, we are usually interested in computing just one of the values of a multiple-
valued function. If we make this choice of value with the concept of continuity in mind,
then we obtain a function that is called a branch of a multiple-valued function. In more
rigorous terms:
“A branch of a multiple-valued function 𝐹(𝑧) is a function 𝑓1 (𝑧) that is continuous
on some domain and that assigns exactly one of the multiple values of 𝐹(𝑧) to
each point 𝑧 in that domain.”
In some books the notion of branch is defined as:
“A branch of a multiple-valued function 𝐹(𝑧) is any single-valued function
𝑓1 (𝑧) that is analytic in some domain at each point 𝑧 of which the value𝑓1 (𝑧)
is one of the values of 𝐹(𝑧) .

Notation: When representing branches of a multiple-valued function 𝐹(𝑧), we will use

lowercase letters with a numerical subscript such as 𝑓1 , 𝑓2 and so on.
 Branches
The requirement that a branch be continuous means that the domain of a branch is
different from the domain of the multiple-valued function.
Example:
The multiple-valued function 𝐹(𝑧) = 𝑧1/2 that assigns to each input 𝑧 the set of two square
roots of 𝑧 is defined for all nonzero complex numbers 𝑧. Even though the principal square
root function 𝑓(𝑧) = 𝑧1/2 does assign exactly one value of 𝐹 to each input 𝑧 (namely, it
assigns to 𝑧 the principal square root of 𝑧), 𝑓(𝑧) is not a branch of 𝐹. The reason for this is
that the principal square root function is not continuous on its domain. 𝑓(𝑧) = 𝑧1/2 is
discontinuous at every point on the negative real axis. Therefore, in order to obtain a branch
of 𝐹(𝑧) = 𝑧1/2 that agrees with the principal square root function, we must restrict the
domain to exclude points on the negative real axis. This gives the function:
𝑓1 𝑧 = 𝑧1/2 = 𝑟𝑒 𝑖𝜃/2 ; −𝜋 < 𝜃 < 𝜋.
The domain of the function 𝑓1 is the set Dom(𝑓1 ) defined by 𝑧 > 0, −𝜋 < arg 𝑧 < 𝜋.
We call the function 𝑓1 𝑧 = 𝑧1/2 , −𝜋 < 𝜃 < 𝜋 the principal branch of 𝐹(𝑧) = 𝑧1/2 .
 Other Branches, Branch Cuts and Branch Points
▪ The choice of branch for any multiple-valued function 𝐹(𝑧) is not unique. We can
make the multiple-valued function a single-valued in other regions of the complex
plane by choosing a different branch for the given function.

▪ A branch cut for a branch 𝑓1 of a multiple-valued function 𝐹 is a portion of a curve

that is excluded from the domain of 𝐹 so that 𝑓1 is continuous on the remaining
points.

▪ Any point that is common to all branch cuts of 𝐹 is called a branch point.
 Example
Although the multiple-valued function 𝐹(𝑧) = 𝑧1/2 is defined for all nonzero complex
numbers ℂ, the principal branch
𝑓1 𝑧 = 𝑟𝑒 𝑖𝜃/2 ; −𝜋 < 𝜃 < 𝜋.
is defined only on the domain |𝑧| > 0, −𝜋 < arg(𝑧) < 𝜋. The non-positive real axis together
with the point 𝑧 = 0, is a branch cut for the principal branch 𝑓1 (𝑧). A different branch of 𝐹(𝑧)
with the same branch cut is given by:
𝑓2 𝑧 = 𝑟𝑒 𝑖𝜃/2 ; 𝜋 < 𝜃 < 3𝜋.
(1+𝑖) −(1+𝑖)
These branches are distinct because for, say, 𝑧 = 𝑖 we have 𝑓1 𝑖 = but 𝑓2 𝑖 = .
2 2
Notice that if we set 𝜑 = 𝜃 − 2𝜋, then the branch 𝑓2 can be
expressed as:
𝑓2 𝑧 = 𝑟𝑒 𝑖(𝜑+2𝜋)/2 = 𝑟𝑒 𝑖𝜑/2 𝑒 𝑖𝜋 = − 𝑟𝑒 𝑖𝜑/2 ; −𝜋 < 𝜑 < 𝜋.
Thus, we have shown that 𝑓2 = −𝑓1 . We can think of these two
branches of 𝐹(𝑧) = 𝑧1/2 as being analogous to the positive and
negative square roots of a positive real number.
 Example
Other branches of 𝐹(𝑧) = 𝑧1/2 can be defined in a similar manner by using any ray
emanating from the origin as a branch cut. In general
𝑓𝑘 𝑧 = 𝑟𝑒 𝑖𝜃/2 ; 𝜃0 < 𝜃 < 𝜃0 + 2𝜋,
defines a branch of 𝐹(𝑧) = 𝑧1/2 . The branch cut for 𝑓𝑘 is the ray 𝑟 ≥ 0, 𝜃 = 𝜃0 , which
includes the origin.
For example,
𝑓3 𝑧 = 𝑟𝑒 𝑖𝜃/2 ; −3𝜋/4 < 𝜃 < 5𝜋/4,
defines a branch of 𝐹(𝑧) = 𝑧1/2 . The branch cut for 𝑓3 is the ray arg(𝑧) = −3𝜋/4
together with the point 𝑧 = 0.

For the present case 𝑧 = 0 is the branch point of 𝐹, since it is on the branch cut of every
branch.
 Practice Questions
Book: A First Course in Complex Analysis with Applications by Dennis
G. Zill and Patrick D. Shanahan.

Chapter: 2

Exercise: 2.6
Q-1 to Q-40
 Analytic Functions
Book: A First Course in Complex Analysis with Applications by
Dennis G. Zill and Patrick D. Shanahan.

• Chapter: 3
• Sections: 3.1
 Derivative of Complex Function
Suppose the complex function 𝑤 = 𝑓 (𝑧), is defined in a neighborhood of a point 𝑧0 . The
complex derivative of 𝑓 at 𝑧0 , denoted by 𝑓 ′ (𝑧0 ), is
′
𝑓 𝑧0 + ∆𝑧 − 𝑓 𝑧0 𝑓 𝑧 − 𝑓 𝑧0
𝑓 (𝑧0 ) = lim = lim , (1)
∆𝑧→0 ∆𝑧 𝑧→𝑧0 𝑧 − 𝑧0
provided this limit exists.
 Example:
Find the derivative of 𝑓(𝑧) = 𝑧 2 − 5𝑧.
Solution:
We solve this by definition. Since we are going to compute the derivative of 𝑓 at any
point, we replace 𝑧0 in (1) by the symbol 𝑧 and first compute:
𝑓 𝑧 + Δ𝑧 − 𝑓 𝑧 = 𝑧 + Δ𝑧 2 − 5 𝑧 + Δ𝑧 − 𝑧 2 − 5𝑧
= 𝑧 2 + 2𝑧Δ𝑧 + Δ𝑧 2 − 5𝑧 − 5Δ𝑧 − 𝑧 2 + 5𝑧
= 2𝑧Δ𝑧 + Δ𝑧 2 − 5Δ𝑧.
So that (1) takes the form:
2
′
2𝑧Δ𝑧 + Δ𝑧 − 5Δ𝑧
𝑓 (𝑧) = lim = lim 2𝑧 + Δ𝑧 − 5 = 2𝑧 − 5.
∆𝑧→0 ∆𝑧 ∆𝑧→0
Thus, the required limit is 𝑓(𝑧) = 2𝑧 − 5.
Alternatively, we notice that 𝑓(𝑧) = 𝑧 2 − 5𝑧 is a complex polynomial function and is
differentiable therefore by using rules of differentiation we have:
′
𝑑 2
𝑓 (𝑧) = (𝑧 − 5𝑧) = 2𝑧 − 5.
𝑑𝑧