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How to prove $\langle x,y\rangle=ax_1\bar{y}_1+bx_2\bar{y}_2$ defines an inner product of $\Bbb{C^2}\iff$$a,b\in\Bbb{R^+}$

$\Rightarrow$ if $\langle x,y\rangle=ax_1\bar{y}_1+bx_2\bar{y}_2$ defines an inner product of $\Bbb{C^2}$, then $\langle x,y\rangle=\overline{\langle y,x\rangle}=\overline{ay_1\bar{x}_1+by_2\bar{x}_2}=\bar{a}x_1\bar{y}_1+\bar{b}x_2\bar{y}_2$. Therefore the imaginary part of $a,b$ has constant $0$, so $a,b\in \Bbb{R^+}$

$\Leftarrow$if $a,b\in\Bbb{R^+}$, we can already prove the symmetry from above. For definite positive, however, how do you know if $x\cdot\bar{x}$ is positive or negative and how do you know $\langle x,x\rangle=0\implies v=0$?. The same question goes for linearity.

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    $\begingroup$ HINT: If $a, b\notin R^+$, the only thing that can go wrong is the property $\langle x,x\rangle >0$ if $x\ne 0$. It is there that you should work $\endgroup$ – Giuseppe Negro Apr 7 '16 at 22:04
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In general if $x \in \mathbb{C}$, then there are real numbers $c,d \in \mathbb{R}$ such that $x = c+id$.

To find the answer to your first two questions, take a look at $x\overline x = (c+id)\overline{(c+id)}$.

For linearity in the first variable notice that both addition and multiplication by a scalar is done coordinate by coordinate and use that complex conjugation is linear.

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