Follow

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.

Sign in to participate in the conversation
Mathstodon

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!