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When calculating the inner product$^1$ of two complex vectors $u$ and $v$, why is the complex conjugate of $v$ used? Why not just compute the inner product as with real vectors?

1:Where the inner product of two vectors is defined as the summation of the product of corresponding elements.

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    A requirement is $\langle z,z \rangle \geq 0$. For $z = i$, without cnjugates you get \langle i, i \rangle = -1 < 0.2012-05-26

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A quick answer is that if $z$ is a complex vector, $\sum_{k=1}^n z_k^2$ will not be real. In this case we do not have $\|z\|^2= \langle z, z\rangle.$

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    @ncmathsadist One last question: If vector $a$ is at $70$ degrees and vector $b$ is at $80$ degrees, the angle between them is $10$. However the complex conjugate of $b$ is at $80+180 = 260$ degrees. Therefore the angle between vector $a$ and the complex conjugate of be vector $b$ is $260-70 = 190$. Therefore won't using the dot product of the *negative conjugate* give you a "wrong" angle? (for example in this case $190$ instead of $10$). I've added 180 degrees to find the complex conjugate because $i$ determines the y axis and multiplying $i$ by $-1$ would rotate the vector 180 degrees.2012-05-28