20-May-25
BELLS UNIVERSITY OF TECHNOLOGY,
OTA, OGUN STATE, NIGERIA
GET 210
ENGINEERING MATHEMATICS II
ENGR. MUBARAK O. ASAFA
ENGINEERING MATHEMATICS
DIFFERENTIATION AND
INTEGRATION OF COMPLEX
FUNCTIONS
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Differentiation of a Complex Function
Introduction.
• Differentiating function of a single real variable, 𝑦 = 𝑓(𝑥), the
derivative of 𝑦 with respect to 𝑥 can be defined as the limiting
value of
as 𝛿𝑥 tends to zero.
𝑦 = 𝑓(𝑥)
i.e.
𝛿𝑦 = 𝑓 𝑥 + 𝛿𝑥 − 𝑓(𝑥)
= lim
( )
→
Differentiation of a Complex Function
For a complex variable, 𝑤 = 𝑓(𝑧), the derivative of 𝑤 with respect
to 𝑧 can similarly be defined as the limiting value of
( )
i.e.
as 𝛿𝑧 tends to zero.
If 𝑃 and 𝑄 in the z-plane map onto 𝑃′ and 𝑄′ in the 𝑤-plane, then
𝑃 𝑄 = 𝛿𝑤 = 𝑤 + 𝛿𝑤 − 𝑤 = 𝑓 𝑧 + 𝛿𝑧 − 𝑓(𝑧 )
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Differentiation of a Complex Function
Therefore, the derivative of 𝑤 at 𝑃′(𝑧 = 𝑧 ) is the limiting value of
𝛿𝑤 as 𝛿𝑧 → 0,
i.e.
= lim
(
)
→
= lim
→
If this limiting value exists, the function 𝑓(𝑧) is said to be
differentiable at 𝑷.
Also, if 𝑤 = 𝑓(𝑧) and 𝑓 𝑧 has a limit for all points 𝑧 within a given
region for which 𝑤 = 𝑓(𝑧) is defined, then 𝑓(𝑧) is said to be
differentiable in that region.
From this, it follows that the limit exists whatever the path of
approach from 𝑄(𝑧 = 𝑧 + 𝛿𝑧) to 𝑃(𝑧 = 𝑧 ).
Differentiation of a Complex Function
Regular Function
Introduction
• A function 𝑤 = 𝑓(𝑧) is said to be regular (or analytic) at a point
𝑧 = 𝑧 , if it is defined and single-valued, and has a derivative at
every point at and around 𝑧 .
• Points in a region where 𝑓(𝑧) ceases to be regular are called
singular points, or singularities.
• A function of a complex variable that is regular over the entire
finite complex plane is called an entire function.
• Examples of entire functions are polynomials, 𝑒 , sin 𝑧 and cos 𝑧.
• In cases where a derivative exists, the usual rules of differentiation
apply.
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Differentiation of a Complex Function
Regular Function
Example 1: Find the derivative of 𝑤 = 𝑧 .
From the first principle,
Given:
𝑤=𝑧
∴ 𝑤 + 𝛿𝑤 = 𝑧 + 𝛿𝑧
= 𝑧 + 2𝑧𝛿𝑧 + 𝛿𝑧
∴ 𝛿𝑤 = 2𝑧𝛿𝑧 + 𝛿𝑧
∴
= 2𝑧 + 𝛿𝑧
∴
= lim 2𝑧 + 𝛿𝑧 = 2𝑧
→
Differentiation of a Complex Function
Example 2: Find the derivative of 𝑤 = 𝑧𝑧̅ where 𝑧 = 𝑥 + 𝑗𝑦 and 𝑧̅ =
𝑥 − 𝑗𝑦.
Given: 𝑤 = 𝑧𝑧̅
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Differentiation of a Complex Function
The next step is to reduce 𝛿𝑧 to zero. But 𝛿𝑧 consists of 𝛿𝑥 + 𝑗𝛿𝑦
and so reducing 𝛿𝑧 to zero can be done by:
(a) Let 𝛿𝑦 → 0 and afterwards let 𝛿𝑥 → 0 as indicated below
=2𝑥
Differentiation of a Complex Function
(b) Let 𝛿𝑥 → 0 and afterwards let 𝛿𝑦 → 0 as indicated below
Then
d𝑤
= lim −𝑗2𝑦 − 𝑗𝛿𝑦 = −𝑗2𝑦
→
d𝑧
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Differentiation of a Complex Function
Note:
• In the first case,
= 2𝑥 while in the second case
= −𝑗2𝑦.
Which are not the same for all values of 𝒙 and 𝒚, except with
𝑥 = 𝑦 = 0.
• Therefore 𝑤 = 𝑧𝑧̅ is a function that is differentiable at a single
point (namely the origin) but nowhere else.
• Consequently, the function is not regular at that point because
to be so it needs to be differentiable not only at the point but
also at points in a region surrounding it.
• Hence the need to test whether or not the function 𝑤 = 𝑓(𝑧) is
or is not regular at all points of a region.
• This useful tool is provided by the Cauchy–Riemann equations.
Differentiation of a Complex Function
Cauchy–Riemann Equations
As in the previous example. If 𝑤 = 𝑓 𝑧 = 𝑢 + 𝑗𝑣, lets establish the
conditions for 𝑤 = 𝑓(𝑧) to have a derivative at a given point 𝑧 = 𝑧 .
𝑤 = 𝑢 + 𝑗𝑣 ∴ 𝛿𝑤 = 𝛿𝑢 + 𝑗𝛿𝑣;
𝑧 = 𝑥 + 𝑗𝑦 ∴ 𝛿𝑧 = 𝛿𝑥 + 𝑗𝛿𝑦
Then 𝑓 𝑧 =
= lim
= lim
→
→
→
(1)
regardless of how 𝛿𝑥 and 𝛿𝑦 tend to zero
(a)
Let 𝛿𝑥 → 0, followed by 𝛿𝑦 → 0
𝑓 𝑧 =
= lim
= lim
−𝑗
→
→
=
−𝑗
(2)
We use the ‘partial’ notation since 𝑢 and 𝑣 are functions of
both 𝑥 and 𝑦.
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Differentiation of a Complex Function
(b)
Cauchy–Riemann Equations
Let 𝛿𝑦 → 0, followed by 𝛿𝑥 → 0.
𝑓 𝑧 =
= lim
= lim
+𝑗
→
→
=
+𝑗
(3)
If the results (2) and (3) are to have the same value for 𝑓 𝑧
irrespective of the path chosen for 𝛿𝑧 to tend to zero, then
+𝑗
=
−𝑗
Equating real and imaginary parts, this gives
=
and
=−
These are the Cauchy–Riemann equations.
Differentiation of a Complex Function
Cauchy–Riemann Equations
Summarily:
A necessary condition for 𝑤 = 𝑓 𝑧 = 𝑢 + 𝑗𝑣 to be regular at 𝑧 = 𝑧
is that 𝑢, 𝑣 and their partial derivatives are continuous and that in
the neighbourhood of 𝑧 = 𝑧 .
=
and
=−
Earlier, it was stated that where a function fails to be regular, a
singular point, or singularity occurs.
For example, where 𝑤 = 𝑓 𝑧
Cauchy–Riemann test fails.
is not continuous or where the
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Differentiation of a Complex Function
Harmonic Functions
If a function of two real variables 𝑓(𝑥, 𝑦) satisfies Laplace’s equation
( , )
( , )
+
=0
then 𝑓(𝑥, 𝑦) is called harmonic function.
To demonstrate that the real and imaginary parts of an analytic
function are both harmonic:
Let 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑗𝑣(𝑥, 𝑦) be an analytic function in some region
of the z-plane.
Because 𝑓(𝑧) is analytic the Cauchy–Riemann equations hold true.
That is
and
=
=−
Differentiation of a Complex Function
Harmonic Functions
Differentiating the first with respect to 𝑥 and the second with respect
to 𝑦 shows gives:
=
and
and so
+
=−
=−
=−
=0
By a similar reasoning
−
and
=
and so
+
=
=
=−
=0
The functions 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are called conjugate functions. In
addition, the curves 𝑢 = constant, 𝑣 = constant are orthogonal.
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Differentiation of a Complex Function
Harmonic Functions
Example 1:
Show that the real and imaginary parts of the function defined by
𝑓(𝑧) = 𝑧 are harmonic.
Differentiation of a Complex Function
Harmonic Functions
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Differentiation of a Complex Function
Harmonic Functions
Example 2
Show that 𝑢 𝑥, 𝑦 = 𝑥3𝑦 − 𝑦3𝑥 is an harmonic function and find the
function 𝑣(𝑥, 𝑦) that ensures that 𝑓 𝑧 = 𝑢 𝑥, 𝑢 + 𝑗𝑣(𝑥, 𝑦) is
analytic. That is, find the function 𝑣(𝑥, 𝑦) that is conjugate to
𝑢(𝑥, 𝑦).
If 𝑓 𝑧 = 𝑢 𝑥, 𝑢 + 𝑗𝑣(𝑥, 𝑦) is analytic then 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦)
satisfy the Cauchy-Riemann equation.
Differentiation of a Complex Function
Example 2
Harmonic Functions
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Differentiation of a Complex Function
Harmonic Functions
Example 2
Now, differentiating this expression with respect to 𝑥 gives
Differentiation of a Complex Function
Example 2
Harmonic Functions
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Differentiation of a Complex Function
Harmonic Functions
Example 3
Given 𝑢(𝑥, 𝑦) = 𝑒 cos 𝑦 , show that 𝑢(𝑥, 𝑦) is an harmonic
function and find the function 𝑣(𝑥, 𝑦) that ensures that 𝑓(𝑧) =
𝑢(𝑥, 𝑦) + 𝑗𝑣(𝑥, 𝑦) is analytic. That is, find the function 𝑣(𝑥, 𝑦) that is
conjugate to 𝑢(𝑥, 𝑦).
that is 𝑢(𝑥, 𝑦) is harmonic
Differentiation of a Complex Function
Example 3
Harmonic Functions
By the Cauchy–Riemann equation
=
Integrating with respect to 𝑦 gives 𝑣 = 𝑒
= −𝑒
cos 𝑦.
sin 𝑦 + 𝑎(𝑥).
Differentiating this with respect to 𝑥 gives
=𝑒
sin 𝑦 + 𝑎′(𝑥).
Now, by the other Cauchy–Riemann equation
𝑒
=−
=
sin 𝑦, so that 𝑎′(𝑥) = 0 giving 𝑎(𝑥) = 𝐶.
Therefore, 𝑣 = −𝑒
sin 𝑦 + 𝐶.
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Complex Integration
Introduction.
Consider 𝑧 as a function of two independent variables 𝑥 and 𝑦, i.e.
𝑧 = 𝑥 + 𝑗𝑦.
𝑤 = 𝑓 𝑧 = 𝑢 + 𝑗𝑣
where 𝑢 and 𝑣 are functions of 𝑥 and 𝑦.
Also, 𝑑𝑧 = 𝑑𝑥 + 𝑗𝑑𝑦 and 𝑑𝑤 = 𝑑𝑢 + 𝑗𝑑𝑣
∴
𝑤𝑑𝑧 =
=
∴
𝑓 𝑧 𝑑𝑧 =
(𝑢 + 𝑗𝑣)(𝑑𝑥 + 𝑗𝑑𝑣)
𝑢𝑑𝑥 − 𝑣𝑑𝑦 + 𝑗(𝑣𝑑𝑥 + 𝑢𝑑𝑦)
𝑓 𝑧 𝑑𝑧 =
𝑢𝑑𝑥 − 𝑣𝑑𝑦 + 𝑗 (𝑣𝑑𝑥 + 𝑢𝑑𝑦)
Complex Integration
Introduction.
That is, the integral reduces to two real-variable integrals
∫ 𝑢𝑑𝑥 − 𝑣𝑑𝑦 and ∫(𝑣𝑑𝑥 + 𝑢𝑑𝑦)
Note that each of these two integrals is of the general form
∫(𝑃𝑑𝑥 + 𝑄𝑑𝑦) which leads us into contour integration.
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Complex Integration
Introduction.
Contour Integration – Line Integrals in the z-plane
 If 𝑧 moves along the curve 𝑐 in the 𝑧-plane and at each position, 𝑧
has a function of 𝑧 associated with it i.e. 𝑓 𝑧 .
 Summing up 𝑓(𝑧) for all such points between 𝐴 and 𝐵 means that
we are evaluating a line integral in the z-plane between 𝐴(𝑧 = 𝑧 )
and B (𝑧 = 𝑧 ) along the curve 𝑐,
 i.e. we are evaluating ∫ 𝑓 𝑧 𝑑𝑧 where 𝑐 is the particular path
joining 𝐴 to 𝐵.
 The evaluation of line integrals in the complex plane is known as
contour integration.
Let us see how it works in practice.
Complex Integration
Example
Evaluate the integral ∫ 𝑓 𝑧 𝑑𝑧, where 𝑓 𝑧 = 𝑧 − 𝑗 and 𝑐 is the
straight line joining 𝐴(𝑧 = 0) to 𝐵(𝑧 = 1 + 𝑗2).
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Complex Integration
Example
Generally, the value of a line integral depends on the path of
integration between the end points, but, the line integral
∫(𝑃𝑑𝑥 + 𝑄𝑑𝑦) is independent of the path of integration in a
simply connected region if
𝝏𝑷
𝝏𝑸
=
throughout the region.
𝝏𝒚
𝝏𝒙
Complex Integration
Example
Therefore, in this example, the value of the line integral is
independent of the path of integration.
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Complex Integration
Cauchy’s Theorem
If 𝑤 = 𝑓(𝑧) where, as usual, 𝑤 = 𝑢 + 𝑗𝑣 and 𝑧 = 𝑥 + 𝑗𝑦, then 𝑑𝑧 =
𝑑𝑥 + 𝑗𝑑𝑦 and
𝑓 𝑧 𝑑𝑧 =
=
(𝑢 + 𝑗𝑣)(𝑑𝑥 + 𝑗𝑑𝑣)
𝑢𝑑𝑥 − 𝑣𝑑𝑦 + 𝑗 (𝑣𝑑𝑥 + 𝑢𝑑𝑦)
If 𝑐 is a closed curve as the path of integration, then
𝑓 𝑧 𝑑𝑧 =
𝑢𝑑𝑥 − 𝑣𝑑𝑦 + 𝑗
(𝑣𝑑𝑥 + 𝑢𝑑𝑦)
Complex Integration
Cauchy’s Theorem
Applying Green’s theorem to each of the two integrals on the righthand side in turn, we have:
(a)
with the first integral:
𝑢𝑑𝑥 − 𝑣𝑑𝑦 =
−
𝜕𝑣 𝜕𝑢
−
𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
where 𝑆 is the region enclosed by the curve 𝑐.
Also, if 𝑓(𝑧) is regular at every point within and on 𝑐, then
the Cauchy–Riemann equations give
=−
∴∮
and therefore −
𝑢𝑑𝑥 − 𝑣𝑑𝑦 = 0
−
=0
(1)
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Complex Integration
(b)
Cauchy’s Theorem
Similarly, with the second integral, we have
𝜕𝑢 𝜕𝑣
𝑣𝑑𝑥 + 𝑢𝑑𝑦 =
−
𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
Again, if 𝑓(𝑧) is regular at every point within and on 𝑐, then
the Cauchy–Riemann equations give
=
and therefore
∴∮
𝑣𝑑𝑥 − 𝑢𝑑𝑦 = 0
−
=0
(2)
Combining the two results (1) and (2) we have the following
result.
If 𝑓(𝑧) is regular at every point within and on a closed curve
𝑐, then ∮ 𝑓 𝑧 𝑑𝑧 = 0. This is known as Cauchy’s Theorem.
Complex Integration
Example 1
Cauchy’s Theorem
Verify Cauchy’s theorem by evaluating the integral ∮ 𝑓 𝑧 𝑑𝑧
where 𝑓 𝑧 = 𝑧 around the square formed by joining the points 𝑧 =
1, 𝑧 = 2, 𝑧 = 2 + 𝑗, 𝑧 = 1 + 𝑗.
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Complex Integration
Cauchy’s Theorem
Example 1
We now take each of the sides in turn.
Complex Integration
Example 1
Cauchy’s Theorem
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Complex Integration
Cauchy’s Theorem
Example 2
A region in the z-plane has a boundary 𝑐 consisting of:
(a)
𝑂𝐴 joining 𝑧 = 0 to 𝑧 = 2
(b)
𝐴𝐵 a quadrant of the circle 𝑧 = 2 from 𝑧 = 2 to 𝑧 = 𝑗2
(c)
𝐵𝑂 joining 𝑧 = 𝑗2 to 𝑧 = 0.
Verify Cauchy’s theorem by evaluating the integral ∫ 𝑧 + 1 𝑑𝑧
(1)
along the arc from 𝐴 to B
(2)
along 𝐵𝑂 and 𝑂𝐴.
Complex Integration
Cauchy’s Theorem
Example 2
The general expression for ∫ 𝑓 𝑧 𝑑𝑧 is given below:
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Complex Integration
Example 2
Cauchy’s Theorem
Complex Integration
Example 2
(2)
Cauchy’s Theorem
Along 𝐵𝑂 and 𝑂𝐴,
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Complex Integration
Example 2
Cauchy’s Theorem
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