Difference of two functions sharing the same poles pole-free?

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SUMMARY

The discussion centers on the mathematical question of whether the difference between two functions, f(z) and g(z), that share the same poles is pole-free. It is established that this is not true, as demonstrated by the counter-example of f(z) = 1/x and g(z) = 1/(2x), both having a pole at zero. The result of their difference, f(z) - g(z) = 1/(2x), retains the pole at zero, confirming that the difference can indeed have poles. Additionally, the case where f(z) = a*g(z) (with a being a constant) further illustrates that f(z) - g(z) shares poles with f(z).

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If f(z) and g(z) share all the same poles, is f(z)-g(z) pole-free? I feel like this would be true, but I can't really come up with a proof for it.
 
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No, consider 1/x and 1/(2x). These both have poles at zero.

But 1/x - 1/(2x)=1/(2x) which also has a pole at zero.

Or did I misunderstood the question?
 
Certainly not. Here's the simplest counter-example: f = a*g where a is a constant, so f-g has the same poles as f. More complicated examples exist as well.
 

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