Purely Real-valued Analytic Functions?

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SUMMARY

A purely real-valued function cannot be analytic unless it is constant. This conclusion arises from the application of the Cauchy-Riemann conditions, which state that for a function to be analytic, it must satisfy the equations ux=vy and vx=-uy. In the case of real-valued functions, the imaginary part v is zero, leading to the conclusion that the real part u must also be constant to satisfy the conditions. Therefore, only constant functions can be classified as purely real analytic functions.

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Homework Statement


Can a function which is purely real-valued be analytic? Describe the behavior of such functions?


Homework Equations


The Cauchy-Riemann conditions
ux=vy, vx=-uy

The Attempt at a Solution



I can't think of any pure real-valued equations off the top of my head which satisfy the CR conditions. However, could any function like sin(x), cos(x), e^x, which is infinitely differentiable be considered analytic? Doesn't infinite differentiability imply that the CR conditions are already satisfied?
 
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jtleafs33 said:

Homework Statement


Can a function which is purely real-valued be analytic? Describe the behavior of such functions?


Homework Equations


The Cauchy-Riemann conditions
ux=vy, vx=-uy

The Attempt at a Solution



I can't think of any pure real-valued equations off the top of my head which satisfy the CR conditions. However, could any function like sin(x), cos(x), e^x, which is infinitely differentiable be considered analytic? Doesn't infinite differentiability imply that the CR conditions are already satisfied?

No. If your function is real valued then v=0. What does that tell you about u?
 
If v=0, then vx=vy=0. So then, u must be constant, so that it's derivative will also be zero, and thus satisfy the CR conditions?
 
jtleafs33 said:
If v=0, then vx=vy=0. So then, u must be constant, so that it's derivative will also be zero, and thus satisfy the CR conditions?

Yes, that's it. The only purely real analytic functions are constant.
 

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