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Show that f'(z) does not exist at any point z for the following functions: (a) f(z) = z - i·z, (b) f(z) = Im(z), (c) f(z) = e^x·e^-iy.

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Final answer:

The first function has a discontinuity, the second does not satisfy the Cauchy-Riemann equations, and the third introduces a divergence, all of which prevent these functions from having a complex derivative at any point z.

Step-by-step explanation:

To show that f'(z) does not exist at any point z for the given functions, we can use the Cauchy-Riemann equations, which are necessary conditions for a function to be differentiable in the complex sense.

Function (a): f(z) = z - i·z

Firstly, we represent f(z) in terms of real and imaginary parts as f(z) = (1-i)z. The function is not continuous as it has a discontinuity caused by the subtraction of the imaginary unit times the input z. Since complex differentiability implies continuity, this function cannot have a derivative at any point z.

Function (b): f(z) = Im(z)

For the function f(z) = Im(z), it only returns the imaginary part of the input, which is not differentiable as a function of a complex variable because it doesn't satisfy the Cauchy-Riemann equations at any point. It is not double-valued, but rather it does not have a complex derivative because it does not influence the real part of z.

Function (c): f(z) = e^x·e^-iy

The last function, f(z) = e^x·e^-iy, can be written as e^x (cos(y) - i sin(y)) using Euler's formula. While the real parts of this function satisfy the Cauchy-Riemann equations, the imaginary parts involve e^-iy, introducing a divergence when considering y to infinity, thus also violating the necessary conditions for differentiability in the complex plane.

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