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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