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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)ℎ or simply 𝑓(𝑧+ℎ)−𝑓(𝑧)ℎ

where take the limit as ℎ→0.

Further examples: 𝑓(𝑧)=𝑧. Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)ℎ= (𝑧0+ℎ)−𝑧0ℎ= ℎℎ=1→1 as ℎ→0

So 𝑓′(𝑧)=1 for all 𝑧∈ℂ.

More examples: 𝑓(𝑧)=𝑧². Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)ℎ= (𝑧0+ℎ)2−𝑧02ℎ= 2𝑧0ℎ+ℎ2ℎ=2𝑧0+ℎ→2𝑧0 as ℎ→0

Thus 𝑓′(𝑧)=2𝑧 for all 𝑧∈ℂ.

Another examples: 𝑓(𝑧)=𝑧^𝑛. Then

𝑓(𝑧0+ℎ)−𝑓(𝑧0)ℎ= (𝑧0+ℎ)𝑛−𝑧0𝑛ℎ=(𝑧0𝑛+𝑛ℎ𝑧0𝑛-1+𝑛(𝑛-1) 2ℎ2𝑧0𝑛-2+⋯+ℎ𝑛)−𝑧0𝑛ℎ
=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)=𝑛𝑧^𝑛-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=ℎ_𝑥+𝑖ℎ_𝑦. Then

  𝑓(𝑧+ℎ)−𝑓(𝑧)ℎ= (𝑥+ℎ𝑥)−𝑥ℎ= ℎ𝑥ℎ= Re ℎℎ

  Does 𝑓(𝑧) have a limit as ℎ→0?

  • ℎ→0 along real axis: Then ℎ=ℎ_𝑥+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⋅ℎ_y, so Re ℎ=0, and thus the quotient evaluates to 0, and the limit equals 0.
  • ℎ_𝑛=𝑖^𝑛𝑛, then Re ℎ_𝑛 ℎ_𝑛=Re 𝑖_𝑛 𝑖_𝑛={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)ℎ= (z+ℎ)−zℎ= ℎℎ
  • If ℎ∈ℝ then ℎℎ=1→1 as ℎ→0
  • If ℎ∈𝑖ℝ then ℎℎ=−1→−1 as ℎ→0

  Thus ℎℎ 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)ℎ= |𝑧+ℎ|2−|𝑧|2ℎ= (𝑧+ℎ)(𝑧+ℎ)−|𝑧|2ℎ= |𝑧|2+𝑧ℎ+ℎ𝑧+ℎℎ−|𝑧|2ℎ= 𝑧+ℎ+𝑧⋅ℎℎ

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