Having encountered this one again recently,

The length of \((i,1)\) in \(\mathbb{C}^2\) is \(\sqrt{2}\) not 1 or 0. You must use the lengths, not the values, when applying Pythagoras.

@kimreece Instead of recalling Pythagoras when dealing with complex vectors, I instead recall it as \(\sqrt{\mathbf v^\ast\cdot\mathbf v}\), which also happens to work in other dimensions, not just 2D.

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𝓙. 𝓜.@tpfto@mathstodon.xyz@kimreece Instead of recalling Pythagoras when dealing with complex vectors, I instead recall it as \(\sqrt{\mathbf v^\ast\cdot\mathbf v}\), which also happens to work in other dimensions, not just 2D.