Proof verification: If $f$ is holomorphic on an open set and if $Re(f)$ is constant, then $f$ is constant.

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Suppose $Re(f)$ is constant. I am trying to prove that $f$ itself is constant. To do so, I want to show that for every $z_0$in our open set, $f'(z_0) = 0.$ We know that $f'(z_0) = fracpartial fpartial x(z_0).$ This is where I make an iffy step. We can write $f$ as $f(x, y) = u(x, y) + v(x, y)i.$ Then $fracpartial fpartial x = fracpartial upartial x + fracpartial vpartial xi.$ By the Cauchy formulas, we have $fracpartial fpartial x = fracpartial vpartial x - fracpartial upartial yi.$ However, since $Re(f)$ is constant, we know that $fracpartial upartial x = fracpartial upartial y = 0.$ Hence, $fracpartial fpartial x(z_0) = 0.$ Does this work?




complex-analysis proof-verification




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    Yes, that's fine.
    – José Carlos Santos
    Aug 2 at 21:25














up vote
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Suppose $Re(f)$ is constant. I am trying to prove that $f$ itself is constant. To do so, I want to show that for every $z_0$in our open set, $f'(z_0) = 0.$ We know that $f'(z_0) = fracpartial fpartial x(z_0).$ This is where I make an iffy step. We can write $f$ as $f(x, y) = u(x, y) + v(x, y)i.$ Then $fracpartial fpartial x = fracpartial upartial x + fracpartial vpartial xi.$ By the Cauchy formulas, we have $fracpartial fpartial x = fracpartial vpartial x - fracpartial upartial yi.$ However, since $Re(f)$ is constant, we know that $fracpartial upartial x = fracpartial upartial y = 0.$ Hence, $fracpartial fpartial x(z_0) = 0.$ Does this work?




complex-analysis proof-verification




share|cite|improve this question

















  • 2




    Yes, that's fine.
    – José Carlos Santos
    Aug 2 at 21:25












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Suppose $Re(f)$ is constant. I am trying to prove that $f$ itself is constant. To do so, I want to show that for every $z_0$in our open set, $f'(z_0) = 0.$ We know that $f'(z_0) = fracpartial fpartial x(z_0).$ This is where I make an iffy step. We can write $f$ as $f(x, y) = u(x, y) + v(x, y)i.$ Then $fracpartial fpartial x = fracpartial upartial x + fracpartial vpartial xi.$ By the Cauchy formulas, we have $fracpartial fpartial x = fracpartial vpartial x - fracpartial upartial yi.$ However, since $Re(f)$ is constant, we know that $fracpartial upartial x = fracpartial upartial y = 0.$ Hence, $fracpartial fpartial x(z_0) = 0.$ Does this work?




complex-analysis proof-verification




share|cite|improve this question













Suppose $Re(f)$ is constant. I am trying to prove that $f$ itself is constant. To do so, I want to show that for every $z_0$in our open set, $f'(z_0) = 0.$ We know that $f'(z_0) = fracpartial fpartial x(z_0).$ This is where I make an iffy step. We can write $f$ as $f(x, y) = u(x, y) + v(x, y)i.$ Then $fracpartial fpartial x = fracpartial upartial x + fracpartial vpartial xi.$ By the Cauchy formulas, we have $fracpartial fpartial x = fracpartial vpartial x - fracpartial upartial yi.$ However, since $Re(f)$ is constant, we know that $fracpartial upartial x = fracpartial upartial y = 0.$ Hence, $fracpartial fpartial x(z_0) = 0.$ Does this work?





complex-analysis proof-verification





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share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 22:22









Berci

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asked Aug 2 at 21:22









伽罗瓦

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  • 2




    Yes, that's fine.
    – José Carlos Santos
    Aug 2 at 21:25












  • 2




    Yes, that's fine.
    – José Carlos Santos
    Aug 2 at 21:25







2




2




Yes, that's fine.
– José Carlos Santos
Aug 2 at 21:25




Yes, that's fine.
– José Carlos Santos
Aug 2 at 21:25















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