Ivan Shishkin, Rye (1878)

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R\R-automorphisms of C\mathbb{C}

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Solution

Solution by Ancient Tree · EN

Let’s assume that such a R\R-automorphism φ:CC\varphi:\mathbb{C}\rightarrow \mathbb{C} exists. Then, since it is a field homomorphism, it must verify:
φ(a+bi)=φ(a)+φ(bi)=φ(a)+φ(b)φ(i)\varphi(a+bi)=\varphi(a)+\varphi(bi)=\varphi(a)+\varphi(b)\varphi(i)but because it is a R\R-morphism, it fixes R\R, so:
φ(a+bi)=a+bφ(i).\varphi(a+bi)=a+b\varphi(i).So φ\varphi would be entirely determined by φ(i)\varphi(i). But we have that φ(1)=φ(i2)=φ(i)2\varphi(-1)=\varphi(i^{2})=\varphi(i)^{2}, which means that φ(i)\varphi(i) is a solution of the equation X2=1X^{2}=-1. This equation admits only two solutions, ii and i-i.
We conclude that either φ(a+bi)=a+bi\varphi(a+bi)=a+bi (this is the identity map), or φ(a+bi)=abi\varphi(a+bi)=a-bi, which is the complex conjugation (and it is indeed a R\R-automorphism).

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