How is the expansion of the square of |z1+z2| derived?
You can use dot product.
|z1+z2|^{2}
= (z1+z2)·(z1+z2)
= z1^{2} + z1·z2 + z2·z1 + z2^{2}
= z1^{2} + z2^{2} + 2z1·z2
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Attn: z1·z2 = z2·z1
How is the expansion of the square of |z1+z2| derived?
You can use dot product.
|z1+z2|^{2}
= (z1+z2)·(z1+z2)
= z1^{2} + z1·z2 + z2·z1 + z2^{2}
= z1^{2} + z2^{2} + 2z1·z2
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Attn: z1·z2 = z2·z1
The absolute value, or modulus of the number z = a + bi is defined by |z| = √(a² + b²)
|z1 + z2| = |(a1 + a2) + (b1 + b2)| = √[(a1 + a2)² + (b1 + b2)²]
|z1 + z2|² = (a1 + a2)² + (b1 + b2)²
z1 = x1 + i*y1
z2 = x2 + i*y2
z1 + z2 = (x1 + x2) + i*(y1 + y2) = X + iY
X = x1 + x2, Y = y1 + y2
i = sqrt(-1), i^2 = -1
|z1 + z2| = |X + iY| = sqrt(X^2 + (i^2)*Y^2) = sqrt(X^2 - Y^2)
|z1 + z2|^2 = (sqrt(X^2 - Y^2))^2 = X^2 - Y^2
Comments
You can use dot product.
|z_{1}+z_{2}|^{2}
= (z_{1}+z_{2})·(z1+z2)
= z_{1}^{2} + z_{1}·z_{2} + z_{2}·z_{1 }+ z_{2}^{2}
= z_{1}^{2} + z_{2}^{2} + 2z_{1}·z_{2}
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Attn: z_{1}·z_{2} = z_{2}·z_{1}