Math Problem Statement

prove that f(z) =(z-1)/(z+1) is differentiable at every point znot equal to -1 and find f'(z)

Solution

To prove that f(z)=z1z+1f(z) = \frac{z-1}{z+1} is differentiable at every point z1z \neq -1, and to find f(z)f'(z), we proceed as follows:


Step 1: Recall the definition of differentiability

A function f(z)f(z) is differentiable at a point zz if the limit

f(z)=limΔz0f(z+Δz)f(z)Δzf'(z) = \lim_{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z}

exists.


Step 2: Express f(z)f(z) as a quotient

We have f(z)=z1z+1f(z) = \frac{z-1}{z+1}. This is a quotient of two functions:

  • Numerator: g(z)=z1g(z) = z-1, differentiable everywhere.
  • Denominator: h(z)=z+1h(z) = z+1, differentiable everywhere except z=1z = -1, where h(z)=0h(z) = 0.

Since the denominator is nonzero at z1z \neq -1, f(z)f(z) is well-defined and differentiable at such points, provided the quotient rule applies.


Step 3: Use the quotient rule to find f(z)f'(z)

The derivative of a quotient f(z)=g(z)h(z)f(z) = \frac{g(z)}{h(z)} is given by:

f(z)=g(z)h(z)g(z)h(z)(h(z))2.f'(z) = \frac{g'(z)h(z) - g(z)h'(z)}{(h(z))^2}.

Here:

  • g(z)=z1g(z) = z-1, so g(z)=1g'(z) = 1,
  • h(z)=z+1h(z) = z+1, so h(z)=1h'(z) = 1.

Substitute these into the formula:

f(z)=(1)(z+1)(z1)(1)(z+1)2.f'(z) = \frac{(1)(z+1) - (z-1)(1)}{(z+1)^2}.

Simplify the numerator:

f(z)=z+1z+1(z+1)2.f'(z) = \frac{z+1 - z + 1}{(z+1)^2}.

f(z)=2(z+1)2.f'(z) = \frac{2}{(z+1)^2}.


Step 4: Conclude differentiability

The derivative f(z)=2(z+1)2f'(z) = \frac{2}{(z+1)^2} exists for all z1z \neq -1, confirming that f(z)f(z) is differentiable at every point z1z \neq -1.


Final Answer:

  1. f(z)=z1z+1f(z) = \frac{z-1}{z+1} is differentiable at every point z1z \neq -1.
  2. The derivative is:

f(z)=2(z+1)2.f'(z) = \frac{2}{(z+1)^2}.


Would you like a detailed example showing the computation of f(z)f'(z) at a specific zz? Let me know!


5 Related Questions:

  1. What does it mean for a function to be differentiable in the complex plane?
  2. How does the quotient rule apply in complex differentiation?
  3. Can f(z)f(z) be differentiable at z=1z = -1? Why or why not?
  4. How does the derivative f(z)f'(z) behave as z1z \to -1?
  5. What are some real-world applications of differentiable complex functions?

Tip:

In complex analysis, always check if the denominator of a quotient is zero at a given point before concluding differentiability.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Differentiation of Complex Functions
Quotient Rule

Formulas

f'(z) = lim(Δz → 0) [(f(z+Δz) - f(z)) / Δz]
Quotient Rule: f'(z) = [g'(z)h(z) - g(z)h'(z)] / (h(z))^2

Theorems

Definition of Differentiability in Complex Analysis
Quotient Rule for Differentiation

Suitable Grade Level

College-level or Advanced High School (Grades 11-12)