Derivatives of Complex Functions - USM

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Derivatives Differentiation Formulas

Derivatives of Complex Functions

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Introduction

1. The idea for the derivative lies in the desire to computeinstantaneous velocities or slopes of tangent lines.

2. In both cases, we want to know what happens when the

denominator in the difference quotientf (x+h)− f (x)

hgoes to

zero.3. Although the visualization is challenging, if not impossible, we

can investigate the same question for functions that map complexnumbers to complex numbers.

4. After all, the algebra and the idea of a limit translate to C.



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Introduction1. The idea for the derivative lies in the desire to compute

instantaneous velocities or slopes of tangent lines.

2. In both cases, we want to know what happens when the


hgoes to






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instantaneous velocities or slopes of tangent lines.2. In both cases, we want to know what happens when the


hgoes to

zero.

3. Although the visualization is challenging, if not impossible, wecan investigate the same question for functions that map complexnumbers to complex numbers.




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hgoes to

zero.3. Although the visualization is challenging

, if not impossible, wecan investigate the same question for functions that map complexnumbers to complex numbers.




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hgoes to

zero.3. Although the visualization is challenging, if not impossible

, wecan investigate the same question for functions that map complexnumbers to complex numbers.




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hgoes to






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hgoes to






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Definition.

Let f be defined in a neighborhood of the point z0. Then fis called differentiable at z0 if and only if the limit

limz→z0

f (z)− f (z0)z− z0

exists.In this case we set

f ′(z0) := limz→z0

f (z)− f (z0)z− z0

and call it the derivative of f at z0. Other notations for the derivative

at z0 aredfdz

(z0) and Df (z0).



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Definition. Let f be defined in a neighborhood of the point z0.

Then fis called differentiable at z0 if and only if the limit

limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0) and Df (z0).



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Definition. Let f be defined in a neighborhood of the point z0. Then fis called differentiable at z0

if and only if the limit

limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0) and Df (z0).



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Definition. Let f be defined in a neighborhood of the point z0. Then fis called differentiable at z0 if and only if the limit

limz→z0

f (z)− f (z0)z− z0

exists.

In this case we set

f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0) and Df (z0).



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limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0) and Df (z0).



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limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0

and call it the derivative of f at z0.

Other notations for the derivative

at z0 aredfdz

(z0) and Df (z0).



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limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0)

and Df (z0).



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limz→z0

f (z)− f (z0)z− z0


f ′(z0) := limz→z0

f (z)− f (z0)z− z0


at z0 aredfdz

(z0) and Df (z0).



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The function f with an open domain is called differentiable

(orholomorphic or analytic) if and only if it is differentiable at every z0in its domain. The function f is called entire if and only if it isdifferentiable at every z ∈ C. If f is not differentiable at z0, but forevery neighborhood of z0 there is a point so that f is analytic in aneighborhood around that point, then z0 is called a singularity.

It can be proved that f is differentiable if and only if

lim∆z→0

f (z0 +∆z)− f (z0)∆z

exists. In this case, the above limit is the

derivative.

With ∆w := f (z0 +∆z)− f (z0) we also writedwdz

= lim∆z→0

∆w∆z

.



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The function f with an open domain is called differentiable (orholomorphic or analytic)

if and only if it is differentiable at every z0in its domain. The function f is called entire if and only if it isdifferentiable at every z ∈ C. If f is not differentiable at z0, but forevery neighborhood of z0 there is a point so that f is analytic in aneighborhood around that point, then z0 is called a singularity.


lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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The function f with an open domain is called differentiable (orholomorphic or analytic) if and only if it is differentiable at every z0in its domain.

The function f is called entire if and only if it isdifferentiable at every z ∈ C. If f is not differentiable at z0, but forevery neighborhood of z0 there is a point so that f is analytic in aneighborhood around that point, then z0 is called a singularity.


lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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The function f with an open domain is called differentiable (orholomorphic or analytic) if and only if it is differentiable at every z0in its domain. The function f is called entire if and only if it isdifferentiable at every z ∈ C.

If f is not differentiable at z0, but forevery neighborhood of z0 there is a point so that f is analytic in aneighborhood around that point, then z0 is called a singularity.


lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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The function f with an open domain is called differentiable (orholomorphic or analytic) if and only if it is differentiable at every z0in its domain. The function f is called entire if and only if it isdifferentiable at every z ∈ C. If f is not differentiable at z0, but forevery neighborhood of z0 there is a point so that f is analytic in aneighborhood around that point, then z0 is called a singularity.


lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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lim∆z→0

f (z0 +∆z)− f (z0)∆z

exists.

In this case, the above limit is the

derivative.


= lim∆z→0

∆w∆z

.



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lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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lim∆z→0

f (z0 +∆z)− f (z0)∆z


derivative.


= lim∆z→0

∆w∆z

.



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Example.

Compute the derivative of f (z) = z3.

dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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Example. Compute the derivative of f (z) = z3.

dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz

= lim∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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dz3

dz= lim

∆z→0

(z+∆z)3− z3

∆z

= lim∆z→0

z3 +3z2∆z+3z∆z2 +∆z3− z3

∆z

= lim∆z→0

3z2∆z+3z∆z2 +∆z3

∆z

= lim∆z→0

∆z(3z2 +3z∆z+∆z2

)∆z

= lim∆z→0

3z2 +3z∆z+∆z2

= 3z2



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Example.

Show that f (z) = z is not differentiable at any z0 ∈ C.

lim∆z→0

z0 +∆z− z0

∆z= lim

∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z

... which does not exist.



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Example. Show that f (z) = z is not differentiable at any z0 ∈ C.

lim∆z→0

z0 +∆z− z0

∆z= lim

∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z




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lim∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z




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lim∆z→0

z0 +∆z− z0

∆z= lim

∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z




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lim∆z→0

z0 +∆z− z0

∆z= lim

∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z




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lim∆z→0

z0 +∆z− z0

∆z= lim

∆z→0

z0 +∆z− z0

∆z

= lim∆z→0

∆z∆z




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Theorem.

If the complex function f is differentiable at z0, then f iscontinuous at z0. Hence, every differentiable function is continuous.However, not every continuous function is differentiable.

Proof. For continuity at z0 note that

0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).

For the last statement, consider f (z) = z.



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Theorem. If the complex function f is differentiable at z0, then f iscontinuous at z0.

Hence, every differentiable function is continuous.However, not every continuous function is differentiable.


0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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Theorem. If the complex function f is differentiable at z0, then f iscontinuous at z0. Hence, every differentiable function is continuous.

However, not every continuous function is differentiable.


0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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Theorem. If the complex function f is differentiable at z0, then f iscontinuous at z0. Hence, every differentiable function is continuous.However, not every continuous function is differentiable.


0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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Proof.

For continuity at z0 note that

0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0

= limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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0 = limz→z0

(z− z0) limz→z0

f (z)− f (z0)z− z0

= limz→z0

(z− z0)f (z)− f (z0)

z− z0

= limz→z0

f (z)− f (z0).




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Theorem.

Let f and g be differentiable at z0 and let c ∈ C. Then thefunctions f +g, f −g and cf are all differentiable at z0 and

(f +g)′(z0) = f ′(z0)+g′(z0),(f −g)′(z0) = f ′(z0)−g′(z0),

(cf )′(z0) = cf ′(z0).



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Theorem. Let f and g be differentiable at z0 and let c ∈ C.

Then thefunctions f +g, f −g and cf are all differentiable at z0 and

(f +g)′(z0) = f ′(z0)+g′(z0),(f −g)′(z0) = f ′(z0)−g′(z0),

(cf )′(z0) = cf ′(z0).



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Theorem. Let f and g be differentiable at z0 and let c ∈ C. Then thefunctions f +g, f −g and cf are all differentiable at z0

and

(f +g)′(z0) = f ′(z0)+g′(z0),(f −g)′(z0) = f ′(z0)−g′(z0),

(cf )′(z0) = cf ′(z0).



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Theorem. Let f and g be differentiable at z0 and let c ∈ C. Then thefunctions f +g, f −g and cf are all differentiable at z0 and

(f +g)′(z0) = f ′(z0)+g′(z0),

(f −g)′(z0) = f ′(z0)−g′(z0),(cf )′(z0) = cf ′(z0).



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(f +g)′(z0) = f ′(z0)+g′(z0),(f −g)′(z0) = f ′(z0)−g′(z0),

(cf )′(z0) = cf ′(z0).



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(f +g)′(z0) = f ′(z0)+g′(z0),(f −g)′(z0) = f ′(z0)−g′(z0),

(cf )′(z0) = cf ′(z0).



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Proof (addition only).

(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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(f +g)′(z0)

= limz→z0

(f +g)(z)− (f +g)(z0)z− z0

= limz→z0

f (z)+g(z)− f (z0)−g(z0)z− z0

= limz→z0

f (z)− f (z0)+g(z)−g(z0)z− z0

= limz→z0

f (z)− f (z0)z− z0

+ limz→z0

g(z)−g(z0)z− z0

= f ′(z0)+g′(z0)



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Theorem.

Product and Quotient Rule. Let f and g be differentiableat z0. Then fg is differentiable at x with

(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).

Moreover, if g(z0) 6= 0, then the quotientfg

is differentiable at z0 with(fg

)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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Theorem. Product and Quotient Rule. Let f and g be differentiableat z0.

Then fg is differentiable at x with

(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).



)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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Theorem. Product and Quotient Rule. Let f and g be differentiableat z0. Then fg is differentiable at x

with

(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).



)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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Theorem. Product and Quotient Rule. Let f and g be differentiableat z0. Then fg is differentiable at x with

(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).



)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).


is differentiable at z0

with(fg

)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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(fg)′(z0) = f ′(z0)g(z0)+g′(z0)f (z0).



)′(z0) =

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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Proof (quotient rule only).

limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0

f (z)g(z0)−f (z0)g(z)g(z)g(z0)

z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z0)+ f (z0)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z0)−(f (z0)g(z)− f (z0)g(z0)

)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0

= limz→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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limz→z0

fg(z)− f

g(z0)

z− z0

= limz→z0

f (z)g(z) −

f (z0)g(z0)

z− z0= lim

z→z0


z− z0

= limz→z0

1g(z)g(z0)

f (z)g(z0)− f (z0)g(z)z− z0

= limz→z0

1g(z)g(z0)


= limz→z0

1g(z)g(z0)


)z− z0

= limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)



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Proof (quotient rule cont.).

limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)= lim

z→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)=

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)

= limz→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)=

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)= lim

z→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)=

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)= lim

z→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)

=f ′(z0)g(z0)−g′(z0)f (z0)(

g(z0))2 .



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limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)= lim

z→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)=

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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limz→z0

1g(z)g(z0)

(f (z)g(z0)− f (z0)g(z0)

z− z0− f (z0)g(z)− f (z0)g(z0)

z− z0

)= lim

z→z0

1g(z)g(z0)

(f (z)− f (z0)

z− z0g(z0)−

g(z)−g(z0)z− z0

f (z0))

=1(

g(z0))2

(f ′(z0)g(z0)−g′(z0)f (z0)

)=

f ′(z0)g(z0)−g′(z0)f (z0)(g(z0)

)2 .



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Theorem.

Chain Rule. Let f ,g be complex functions and let z0 besuch that g is differentiable at z0 and f is differentiable at g(z0). Thenf ◦g is differentiable at z0 and the derivative is(f ◦g)′(z0) = f ′

(g(z0)

)g′(z0).



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Theorem. Chain Rule.

Let f ,g be complex functions and let z0 besuch that g is differentiable at z0 and f is differentiable at g(z0). Thenf ◦g is differentiable at z0 and the derivative is(f ◦g)′(z0) = f ′

(g(z0)

)g′(z0).



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Theorem. Chain Rule. Let f ,g be complex functions and let z0 besuch that g is differentiable at z0 and f is differentiable at g(z0).

Thenf ◦g is differentiable at z0 and the derivative is(f ◦g)′(z0) = f ′

(g(z0)

)g′(z0).



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Theorem. Chain Rule. Let f ,g be complex functions and let z0 besuch that g is differentiable at z0 and f is differentiable at g(z0). Thenf ◦g is differentiable at z0

and the derivative is(f ◦g)′(z0) = f ′

(g(z0)

)g′(z0).



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Theorem. Chain Rule. Let f ,g be complex functions and let z0 besuch that g is differentiable at z0 and f is differentiable at g(z0). Thenf ◦g is differentiable at z0 and the derivative is(f ◦g)′(z0) = f ′

(g(z0)

)g′(z0).



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Proof (small technicality ignored).

limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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limz→z0

f ◦g(z)− f ◦g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)z− z0

· g(z)−g(z0)g(z)−g(z0)

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· g(z)−g(z0)z− z0

= limz→z0

f(g(z)

)− f

(g(z0)

)g(z)−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= limu→g(z0)

f (u)− f(g(z0)

)u−g(z0)

· limz→z0

g(z)−g(z0)z− z0

= f ′(g(z0)

)g′(z0).



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Example

(derivatives work just like they did in calculus). Find thederivative of f (x) = (iz+4)3 (

z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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Example (derivatives work just like they did in calculus).

Find thederivative of f (x) = (iz+4)3 (

z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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Example (derivatives work just like they did in calculus). Find thederivative of f (x) = (iz+4)3 (

z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2

+(iz+4)3 2(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)

= (iz+4)2 (z2 +1

)(7iz2 +16z+3i

)



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z2 +1)2.

ddz

(iz+4)3 (z2 +1

)2

= 3(iz+4)2 i(z2 +1

)2+(iz+4)3 2

(z2 +1

)2z

= 3i(iz+4)2 (z2 +1

)2+4z(iz+4)3 (

z2 +1)

= (iz+4)2 (z2 +1

)(3iz2 +3i+4iz2 +16z

)= (iz+4)2 (

z2 +1)(

7iz2 +16z+3i)



Documents

Derivatives of Complex Functions - USM