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Let $u,v$ unit vectors in $C^n$ so $$||u+v||=\sqrt{2}$$. Need to prove that $<u,v>=bi$ for any b real number.

Please help me Im not sure if I even open $$||u+v||=\sqrt{2}$$ .

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up vote 3 down vote accepted

Square both sides. You get

$$||u+v||^2 = ||u||^2+||v||^2 + 2Re(<u,v>) = 2 $$

Therefore $Re(<u,v>) = 0.$ Thus $<u,v>$ must be imaginary, if not 0.

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