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$..$ Consider inner product space : $(C, \langle \cdot,\cdot\rangle)$: where for complex numbers $..$ $\langle z_1, z_2 \rangle = \sqrt(z_1 *\overline{z_2}$)

Computing $..$ $\langle 2-3i, 2-3i \rangle = \sqrt((2-3i)(2+3i)) = \sqrt 13$

I am not sure how they got 13 as the answer when i is not known?

Any help is much appreciated!

NOTE: the square root should cover whatever is in parenthesis (not sure how to make that happen)

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  • $\begingroup$ $i$ is the imaginary unit. It has property $i^2=-1$. $\endgroup$ – user105013 Dec 15 '14 at 16:50
  • $\begingroup$ @user105013 oh imaginary unit, fun...XD haha well thanks for the information that helps. Since it is basically the answer if you want to go ahead and answer it i'll accept it as answer! Thanks! $\endgroup$ – Jacob Dec 15 '14 at 16:53
  • $\begingroup$ The square root must be wrong if this is meant to be an inner product. An inner product must be linear in one of the variables (and conjugate-linear in the other) which is impossible with the square root. $\endgroup$ – copper.hat Dec 15 '14 at 17:05
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$i$ is the imaginary unit. It has property $i^2=-1$.

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Remember easier way to see this is that $(Number * its conjugate)^2 = (Real^2) + (Imaginary)^2$

so in your case its $4 + 9 = 13 $.

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    $\begingroup$ thanks for the additional tip! $\endgroup$ – Jacob Dec 15 '14 at 17:28

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