Inner prodcut space for complex numbers including complex conjugation

Question

$..$ 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)

Answer 1

$i$ is the imaginary unit. It has property $i^2=-1$.

Answer 2

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

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