Is Each Transformed Function Analytic in Domain D?

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SUMMARY

The discussion centers on the analyticity of transformed functions derived from an analytic function f0 = u(x,y) + i*v(x,y) in domain D. It is established that f1 = (u^2 - v^2) - 2i*u*v is analytic in domain D, as it is the square of f0, which is analytic. For f2 = (e^u)cos(v) + i*(e^u)sin(v), the conclusion is that it is also analytic due to the composition of continuous and differentiable functions. However, f3 = u - i*v is not analytic because the conjugate of an analytic function is not differentiable everywhere. The Cauchy-Riemann equations are referenced as foundational to these conclusions.

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Homework Statement


Given that f0 = u(x,y) + i*v(x,y) is analytic in domain D, are these functions analytic in domain D?
1. f1 = (u^2 - v^2) - 2i*u*v
2. f2 = (e^u)cos(v) + i*(e^u)sin(v)
3. f3 = u - i*v


Homework Equations


cauchy-riemann equations

The Attempt at a Solution


I kind of have an idea but I want to double check if my reasoning is right.

If f0 is analytic then all of the partial derivatives of u(x,y) and v(x,y) wrt x and y exist and are continuous on domain D.

1. f1 = f0^2. The product of differentiable and continuous function is also differentiable and continuous on domain D. Therefore f1 is analytic on domain D.

2. f2 = e^u * 0.5*(e^-v + e^v) + i * e^u * 0.5(e^-v - e^v)
Unsure. I can't explicity use Cauchy-Riemann to check. But I know that v and u are continuous. If a function g(x) is continuous and differentiable on some domain or set, is h(x) = e^g(x) also? If so, then f2 is analytic.

3. f3 = conjugate(f0). Not differentiable because conjugate(z) is not differentiable everywhere. Therefore not analytic.
 
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I agree with all of those. h(x)=e^g(x) is analytic if g(x) is analytic. It's a composition of analytic functions. If f(z) is analytic and g(z) is analytic then f(g(z)) is analytic.
 

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