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  "body": "\n## Proof\n\nWrite $\\bar a$ for the [[complex conjugate|0BJV24C4SY]] of a scalar $a$. Over $\\R$, $\\bar a = a$.\n\n1. By axioms 1 and 2 of the [[inner product|0NB0AWGJBW]], for $u, v, w \\in V$ and a scalar $a$, we get $\\langle w, u + a v \\rangle = \\overline{\\langle u, w \\rangle + a \\langle v, w \\rangle}$. This equals $\\langle w, u \\rangle + \\bar a \\langle w, v \\rangle$.\n2. Let $a$ be a scalar with $a \\bar a = 1$. By the [[induced norm|0MFJJV97RJ]], axiom 1 and step 1, $\\norm{x + a y}^2 = \\langle x + a y, x + a y \\rangle$.\n3. Expanding, $\\norm{x + a y}^2 = \\langle x, x \\rangle + \\bar a \\langle x, y \\rangle + a \\langle y, x \\rangle + a \\bar a \\langle y, y \\rangle$. Since $a \\bar a = 1$, the last term is $\\norm{y}^2$.\n4. Real case. Over $\\R$, axiom 2 gives $\\langle y, x \\rangle = \\langle x, y \\rangle$. Step 3 with $a = 1$ gives $\\norm{x + y}^2 = \\norm{x}^2 + 2 \\langle x, y \\rangle + \\norm{y}^2$.\n5. Step 3 with $a = -1$ gives $\\norm{x - y}^2 = \\norm{x}^2 - 2 \\langle x, y \\rangle + \\norm{y}^2$. Subtracting from step 4, $\\norm{x + y}^2 - \\norm{x - y}^2 = 4 \\langle x, y \\rangle$.\n6. Dividing by $4$ gives the real identity.\n7. Complex case. Let $a = i^k$ with $k \\in \\set{0, 1, 2, 3}$. Then $\\bar a = (-i)^k$, so $a \\bar a = 1^k = 1$. Step 3 applies.\n8. Multiply step 3 by $a$ and use $a \\bar a = 1$. This gives $a \\norm{x + a y}^2 = a \\bigl( \\norm{x}^2 + \\norm{y}^2 \\bigr) + \\langle x, y \\rangle + a^2 \\langle y, x \\rangle$.\n9. Sum step 8 over $k = 0, 1, 2, 3$. We have $\\sum_{k=0}^{3} i^k = 1 + i - 1 - i = 0$.\n10. Also $\\sum_{k=0}^{3} i^{2k} = 1 - 1 + 1 - 1 = 0$. So only the term $\\langle x, y \\rangle$ survives, four times.\n11. Hence $\\sum_{k=0}^{3} i^k \\norm{x + i^k y}^2 = 4 \\langle x, y \\rangle$. Dividing by $4$ gives the complex identity.\n\n## Idea\n\nExpanding $\\norm{x + a y}^2$ gives $\\norm{x}^2 + \\norm{y}^2$ plus the mixed terms $\\bar a \\langle x, y \\rangle + a \\langle y, x \\rangle$. Weighting by suitable roots of unity and summing cancels everything except $\\langle x, y \\rangle$. Over $\\R$, the weights $\\pm 1$ are enough, since the inner product is symmetric.\n",
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