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    Open Mapping Theorem

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    Let P : C -> R be defined by P(z) = Re z; show that P is an open map but it is not a closed map. ( Hint: Consider the set F = { z : Imz = ( Re z)^-1 and Re z doesn't equal to 0}.) Please explain every step and justify.

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    https://brainmass.com/math/complex-analysis/open-mapping-theorem-60955

    Solution Preview

    Proof:
    First, I show that P is an open map. We consider any open set D in C, I want to show that P(D) is an open set in R. For any x in P(D), we can find some z in D, such that P(z)=x. Since P(z)=Re z, then Re z=x. So z=x+iy for some y. ...

    Solution Summary

    This is a proof that a map is an open map but not closed map.

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