Question

Consider the function f(z)=zRe(z)

1. We have several results and examples illustrating the concept of continuity. In terms of these results and examples, give a mathematically valid explanation as to why f is continuous at every z∈C.

2. Show that f cannot be differentiable at any nonzero complex number.

3. Prove f is differentiable only at z=0.

Answer 1

To show that the function f(z)=zRe(z) is continuous at every z∈C, we can use the concept of continuity, the properties of complex numbers, and the Cauchy-Riemann equations. The function is continuous everywhere in the complex plane, but it is not differentiable at any nonzero complex number except for z=0.

Step-by-step explanation:

To show that the function f(z)=zRe(z) is continuous at every z∈C, we can use the concept of continuity and the properties of complex numbers.

1. Continuity of f(z) at every z∈C:

Since f(z) is a product of two continuous functions, z and Re(z), it is also continuous. The function z is continuous everywhere in the complex plane, and Re(z) is continuous at every point as it is a real-valued function. Therefore, the product of two continuous functions is also continuous.

2. Differentiability of f(z) at any nonzero complex number:

The function f(z)=zRe(z) cannot be differentiable at any nonzero complex number because it fails to satisfy the Cauchy-Riemann equations. By applying the Cauchy-Riemann equations, we can show that the partial derivatives of f(z) with respect to x and y do not exist simultaneously.

3. Differentiability of f(z) only at z=0:

At z=0, f(z) becomes f(0)=0. The function f(z) is differentiable only at z=0 because it satisfies the Cauchy-Riemann equations. The partial derivatives of f(z) with respect to x and y exist and are equal to each other at z=0, indicating differentiability.