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Complex Derivative
 Derivative of a Function
 The Complex Derivative
  Other Forms of the Difference Quotient
 Differentiation Rules
 Differentiability of a Function

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Complex Derivative

Derivative of a Function

Let 𝑓:(𝑎,𝑏)→ℝ be a real-valued function of a real variable, and let 𝑥0∈(𝑎,𝑏). The function 𝑓 is differentiable at 𝑥0 if lim𝑥→𝑥0 𝑓(𝑥)−𝑓(𝑥0)𝑥−𝑥0 exist. If so, we call this limit the derivative of 𝑓 at 𝑥0 and dente it by 𝑓'(𝑥0).

𝑓(𝑥)−𝑓(𝑥0)𝑥−𝑥0 is the slope of the secant line through the points (𝑥0, 𝑓(𝑥0)) and (𝑥, 𝑓(𝑥)). The slope of the secant line changes as 𝑥 approaches 𝑥0. In the limit, the slopes approach the slope of the tangent line to the graph of 𝑓 at 𝑥0.

However, the derivative does not always exist. For exampe, the graph of 𝑓 does not have a tangent line at 𝑥0.

The Complex Derivative

By definition. A complex-valued function 𝑓 of a complex variable is (complex) differentiable at 𝑧0∈domain(𝑓) if lim𝑧→𝑧0 𝑓(𝑧)−𝑓(𝑧0)𝑧−𝑧0 exist.

If this limit exist, it is denoted 𝑓′(𝑧0) or 𝖽𝑓𝖽𝑧(𝑧0), or 𝖽𝖽𝑧𝑓(𝑧)
𝑧=𝑧0
.

Example: 𝑓(𝑧)=𝖼 (a constant function, 𝖼∈ℂ).

Let 𝑧0∈ℂ be arbitrary. Then 𝑓(𝑧)−𝑓(𝑧0)𝑧−𝑧0= 𝖼−𝖼𝑧−𝑧0=0→0 as 𝑧→𝑧0

Thus 𝑓'(𝑧)=0 for all 𝑧∈ℂ.

Other Forms of the Difference Quotient

Instead of using 𝑓(𝑧)−𝑓(𝑧0)𝑧−𝑧0

Also often write as 𝑧=𝑧0+𝗁 (where 𝗁∈ℂ), and the difference quotient becomes

𝑓(𝑧0+ℎ)−𝑓(𝑧0) or simply 𝑓(𝑧+ℎ)−𝑓(𝑧)

where take the limit as ℎ→0.

Further examples: 𝑓(𝑧)=𝑧. Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)= (𝑧0+ℎ)−𝑧0= =1→1 as ℎ→0

So 𝑓′(𝑧)=1 for all 𝑧∈ℂ.

More examples: 𝑓(𝑧)=𝑧2. Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)= (𝑧0+ℎ)2−𝑧02= 2𝑧0ℎ+ℎ2=2𝑧0+ℎ→2𝑧0 as ℎ→0

Thus 𝑓′(𝑧)=2𝑧 for all 𝑧∈ℂ.

Another examples: 𝑓(𝑧)=𝑧𝑛. Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)= (𝑧0+ℎ)𝑛−𝑧0𝑛=(𝑧0𝑛+𝑛ℎ𝑧0𝑛-1+𝑛(𝑛-1) 22𝑧0𝑛-2+⋯+ℎ𝑛)−𝑧0𝑛
=𝑛𝑧0𝑛-1+𝑛(𝑛-1) 2ℎ𝑧0𝑛-2+⋯+ℎ𝑛-1=𝑛𝑧0𝑛-1+ℎ(𝑛(𝑛-1) 2𝑧0𝑛-2+⋯+ℎ𝑛-2)→𝑛𝑧0𝑛-1 as ℎ→0

Thus 𝑓′(𝑧)=𝑛𝑧𝑛-1 for all 𝑧∈ℂ.

Differentiation Rules

By theorem. Suppose 𝑓 and 𝑔 are differentiable at 𝑧, and ℎ is differentiable at 𝑓(𝑧). Let 𝑐∈ℂ. Then

  • (𝑐𝑓)′(𝑧)=𝑐𝑓′(𝑧)
  • (𝑓+𝑔)′(𝑧)=𝑓′(𝑧)+𝑔′(𝑧)
  • (𝑓*𝑔)′(𝑧)=𝑓′(𝑧)𝑔(𝑧)+𝑓(𝑧)𝑔′(𝑧) Product Rule
  • (𝑓𝑔)′(𝑧)= 𝑔(𝑧)𝑓′(𝑧)−𝑓(𝑧)𝑔′(𝑧)(𝑔(𝑧))2, for 𝑔(𝑧)≠0 Quotient Rule
  • (ℎ∘𝑓)′(𝑧)=ℎ′(𝑓(𝑧))𝑓′(𝑧) Chain Rule

Differentiability of a Function

Differentiable example

  • 𝑓(𝑧)=5𝑧3+s𝑧2-𝑧+7 then 𝑓′(𝑧)=5⋅3𝑧2+2⋅2𝑧−1=15𝑧2+4𝑧−1
  • 𝑓(𝑧)=1𝑧 then 𝑓′(𝑧)=𝑧⋅0−1⋅1 𝑧2=−1𝑧2
  • 𝑓(𝑧)=(𝑧2−1)𝑛 then 𝑓′(𝑧)=𝑛(𝑧2−1)𝑛−1⋅2𝑧
  • 𝑓(𝑧)=(𝑧2−1)(3𝑧+4) then 𝑓′(𝑧)=(2𝑧)(3𝑧+4)+(𝑧2−1)⋅3
  • 𝑓(𝑧)=𝑧𝑧2+1 then 𝑓′(𝑧)= (𝑧2+1)−𝑧⋅2𝑧(𝑧2+1)2= 1−𝑧2(1+𝑧2)2

Non-differentiable example

  • Let 𝑓(𝑧)=Re (𝑧). Write 𝑧=𝑥+𝑖𝑦 and ℎ=ℎ𝑥+𝑖ℎ𝑦. Then

    𝑓(𝑧+ℎ)−𝑓(𝑧)= (𝑥+ℎ𝑥)−𝑥= 𝑥= Re

    Does 𝑓(𝑧) have a limit as ℎ→0?

    • ℎ→0 along real axis: Then ℎ=ℎ𝑥+𝑖⋅0 , so Re ℎ=ℎ, and thus the quotient evaluates to 1, and the limit equals 1.
    • ℎ→0 along imaginary axis: Then ℎ=0+𝑖⋅ℎy, so Re ℎ=0, and thus the quotient evaluates to 0, and the limit equals 0.
    • 𝑛=𝑖𝑛𝑛, then Re𝑛 𝑛=Re 𝑖𝑛 𝑖𝑛={1 if 𝑛 is even0 if 𝑛 is odd has no limit as n→∞.

    𝑓 is not differentiable anywhere in ℂ.

  • Let 𝑓(𝑧)=𝑧 then

    𝑓(𝑧+ℎ)−𝑓(𝑧)= (z+)−z=
    • If ℎ∈ℝ then =1→1 as ℎ→0
    • If ℎ∈𝑖ℝ then =−1→−1 as ℎ→0

    Thus does not have a limit as ℎ→0, and 𝑓 is not differentiable anywhere in ℂ.

By Fact. If 𝑓 is differentiable at z0 then 𝑓 is continuos at 𝑧0.

Proof

lim𝑧→𝑧0 (𝑓(𝑧)−𝑓(𝑧0))=lim𝑧→𝑧0( 𝑓(𝑧)−𝑓(𝑧0)𝑧−𝑧0⋅(𝑧−𝑧0))=𝑓′(𝑧0)⋅0=0

Note however that a function can be continuous without being differentiable.

By definition. A function 𝑓 is analytic in an open set 𝑈⊂ℂ if 𝑓 is (complex) differentiable at each point 𝑧∈𝑈. A function which is analytic in all of ℂ is called an entire function.

Examples:

  • polynomials are analytic in ℂ (hence entire)
  • rational functions 𝑝(𝑧)𝑎(𝑧)are analytic wherever 𝑎(𝑧)≠0
  • 𝑓(𝑧)=𝑧 is not analytic
  • 𝑓(𝑧)=Re z is not analytic

Another examples:

Let 𝑓(𝑧)=|𝑧|2, then

𝑓(𝑧+ℎ)−𝑓(𝑧)= |𝑧+ℎ|2−|𝑧|2= (𝑧+ℎ)(𝑧+)−|𝑧|2= |𝑧|2+𝑧+ℎ𝑧+ℎ−|𝑧|2= 𝑧++𝑧⋅

Thus,

  • If 𝑧≠0 then the limit as ℎ→0 does not exist.
  • If 𝑧=0 then the limit equals 0, thus 𝑓 is differentiable at 0 with 𝑓′(𝑧)=0.
  • 𝑓 is not analytic anywhere
  • Note: 𝑓 is continuous in ℂ

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ID: 190300020 Last Updated: 20/3/2019 Revision: 0

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