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Question 5. Let f: C→C be defined by f(z) = |(Re(z))²(Im(z))]. Show that f satisfies the Cauchy-Riemann equations at z = 0, but is not differentiable at 0.

Question 5. Let f: C→C be defined by f(z) = |(Re(z))²(Im(z))]. Show that f satisfies the Cauchy-Riemann equations at z = 0, but is not differentiable at 0.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 77E
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Question 5. Let f: C → C be defined by ƒ(z) = √|(Re(z))²(Im(z))|. Show that f satisfies the Cauchy-Riemann equations at z = 0, but is not differentiable at 0. Theorem 1.4.8 states conditions under which the Cauchy-Riemann equations holding does imply differentiability. It is a slightly fussy statement, with more conditions on the partial derivatives of the real and imaginary parts of the function. This question demonstrates why, by giving an example of a function which does satisfy the Cauchy- Riemann equations at a point, but is not differentiable at that point. When checking the Cauchy-Riemann equations, observe that you are only verifying them at a particular (rather easy!) point, so there should not be any need to calculate general formulae for the partial derivatives.
Transcribed Image Text:Question 5. Let f: C → C be defined by ƒ(z) = √|(Re(z))²(Im(z))|. Show that f satisfies the Cauchy-Riemann equations at z = 0, but is not differentiable at 0. Theorem 1.4.8 states conditions under which the Cauchy-Riemann equations holding does imply differentiability. It is a slightly fussy statement, with more conditions on the partial derivatives of the real and imaginary parts of the function. This question demonstrates why, by giving an example of a function which does satisfy the Cauchy- Riemann equations at a point, but is not differentiable at that point. When checking the Cauchy-Riemann equations, observe that you are only verifying them at a particular (rather easy!) point, so there should not be any need to calculate general formulae for the partial derivatives.
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