Math Problem Statement

determine the derivative of f(z)= (2z+i)/(z-2i). using the definition f'(z)= lim as h approaches 0 ((f(z+h))-f(z))/h

Solution

To determine the derivative of f(z)=2z+iz2if(z) = \frac{2z + i}{z - 2i} using the definition of the derivative:

f(z)=limh0f(z+h)f(z)h,f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}, follow these steps:


Step 1: Substitute f(z)f(z) and f(z+h)f(z+h) into the definition

  1. f(z)=2z+iz2if(z) = \frac{2z + i}{z - 2i},
  2. f(z+h)=2(z+h)+i(z+h)2i=2z+2h+iz+h2if(z+h) = \frac{2(z+h) + i}{(z+h) - 2i} = \frac{2z + 2h + i}{z + h - 2i}.

The difference quotient becomes: f(z+h)f(z)h=2z+2h+iz+h2i2z+iz2ih.\frac{f(z+h) - f(z)}{h} = \frac{\frac{2z + 2h + i}{z + h - 2i} - \frac{2z + i}{z - 2i}}{h}.


Step 2: Simplify the numerator

Find a common denominator for the two fractions: Numerator: 2z+2h+iz+h2i2z+iz2i=(2z+2h+i)(z2i)(2z+i)(z+h2i)(z+h2i)(z2i).\text{Numerator: } \frac{2z + 2h + i}{z + h - 2i} - \frac{2z + i}{z - 2i} = \frac{(2z + 2h + i)(z - 2i) - (2z + i)(z + h - 2i)}{(z + h - 2i)(z - 2i)}.

Expand each term:

  • (2z+2h+i)(z2i)=2z2+2hz+iz4zi4hi2i2(2z + 2h + i)(z - 2i) = 2z^2 + 2hz + iz - 4zi - 4hi - 2i^2,
  • (2z+i)(z+h2i)=2z2+2hz+iz+2zi+2hi2i2(2z + i)(z + h - 2i) = 2z^2 + 2hz + iz + 2zi + 2hi - 2i^2.

Subtract the two terms: (2z2+2hz+iz4zi4hi2i2)(2z2+2hz+iz+2zi+2hi2i2),(2z^2 + 2hz + iz - 4zi - 4hi - 2i^2) - (2z^2 + 2hz + iz + 2zi + 2hi - 2i^2), =6zi6hi.= -6zi - 6hi.

So the numerator becomes: 6zi6hi(z+h2i)(z2i).\frac{-6zi - 6hi}{(z + h - 2i)(z - 2i)}.


Step 3: Divide by hh

Divide by hh: 6zi6hi(z+h2i)(z2i)h=6zi6hih(z+h2i)(z2i)=6zi(z+h2i)(z2i)6i(z+h2i)(z2i).\frac{\frac{-6zi - 6hi}{(z + h - 2i)(z - 2i)}}{h} = \frac{-6zi - 6hi}{h(z + h - 2i)(z - 2i)} = \frac{-6zi}{(z + h - 2i)(z - 2i)} - \frac{6i}{(z + h - 2i)(z - 2i)}.


Step 4: Take the limit as h0h \to 0

As h0h \to 0, the terms involving hh vanish: 6i(z+h2i)(z2i)6i(z2i)2.\frac{-6i}{(z + h - 2i)(z - 2i)} \to \frac{-6i}{(z - 2i)^2}.

So the derivative is: f(z)=6i(z2i)2.f'(z) = \frac{-6i}{(z - 2i)^2}.


Final Answer:

f(z)=6i(z2i)2.f'(z) = \frac{-6i}{(z - 2i)^2}.

Let me know if you need a detailed explanation for any step or have further questions.


Related Questions:

  1. How do you compute derivatives for other rational functions using the definition?
  2. What role does the denominator's square play in the result?
  3. How does the result differ when the numerator involves quadratic or higher terms?
  4. Can this method be applied to functions with more than one complex variable?
  5. How does this derivative relate to the Cauchy-Riemann equations?

Tip:

When working with complex derivatives, simplify carefully and watch for terms that cancel when applying the limit.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Derivatives
Limits
Rational Functions

Formulas

f'(z) = lim(h → 0) [(f(z+h) - f(z))/h]
Derivative of a rational function: (u/v)' = (u'v - uv')/v^2

Theorems

Definition of Derivative

Suitable Grade Level

Undergraduate Mathematics