analytic proof?

Question

Prove that f(z) = absolute value of z is not analytic where z is a complex number.

If I am to show that this does not verify the cauchy riemann equations, whats the derivative of abs(x+iy)?

Answer 1

You could do this via the definition of derivative. To check the point i, approach from two different ways, say on the line y=i (in the complex plane) and on the imaginary axis (x=0). You should get the derivative to be different. You can then generalize this for complex numbers.

Note: this is similar to showing that the derivative of y=|x| does not exist at x=0.

Answer 2

Here's a very naive take on this. Let's say f(z) = abs(x+iy):
|--: For f(z) >0, the derivative is uniformly 1.
|--: For f(z) =0, the derivative is uniformly 0.
|--: For f(z) <0, the derivative is uniformly -1.

Hence there is a discontinuity at f'(z)=0, just as there is for f'(x)=0.

Answer 3

Man, I hardly remember Complex Analisys, but the the function may not be analytical in the whole plane x+yi, and be in subdomains of C. Then, apart from singularities, the derivative exists in the rest (or parts) of C. ???

proof implies explicit definition. That's maths.

Answer 4

Negating the time period 2(a?b) above calls for the "definition" that 2 vectors are orthogonal even as their dot product is 0. the concept the dot product equivalent to 0 is an similar because the Euclidean idea of perpendicular is known without evidence. The dot product equivalent to 0 extends to any type of dimensions. while, operations utilising the Euclidean idea of perpendicular attitude and Euclidean distance, together with the flow product are in common words valid as a lot as 3 dimensions, which exhibits the mapping of vector orthogonality to Euclidean perpendicular isn't a a million-a million and onto mapping. Translating vector orthogonality to the idea of perpendicular attitude for a Pythagorean evidence calls for the regulation of Cosines that is derived from the Pythagorean theorem (making the argument round).