Why in complex dot product, $\vec{a}\cdot\vec{b}=0$ when $\vec{a}$ and $\vec{b}$ are orthogonal? $\vec{a}\cdot\vec{b}=x_1x_2+y_1y_2+...+x_ny_n=|\vec{a}||\vec{b}|\cos\theta$
Therefore when $\theta$ is $\pi/2$ ,
$\vec{a}\cdot\vec{b}=0$
But for complex inner product:
$\vec{a}\cdot\vec{b}=x_1\bar{y_1}+x_2\bar{y_2}+...+x_n\bar{y_n}$ where $\bar{y}$ is complex conjugate. How can we derive $\vec{a}$ and $\vec{b}$ are orthogonal when $\vec{a}\cdot\vec{b}=0$?
 A: For the real vector space $\mathbb{R}^n$ we use the dot product to define the (cosine of) the angle between two vectors. That definition matches the geometric definition of angle in the plane the vectors determine. Then "orthogonal" means "perpendicular to" (for nonzero vectors), so it matches our geometric idea of orthogonality.
In the complex vector spaces  $\mathbb{C}^n$ we define the dot product using complex conjugation so that it is always nonnegative and has a nice relationship to the norm (length). We carry over the definition of orthogonal, since it is very useful. But it no longer means "perpendicular" in the usual geometric sense.
A: The definition of two vectors being orthogonal is that their dot product is zero. So you don't need to derive that $a \cdot b = 0$ implies that they are orthogonal, that is literally what orthogonal means.
The reason why $a \cdot b$ is defined in that way for complex vectors is so that their norm is always positive: observe that if $a$ is a complex vector, then $a_1 \cdot a_1 + \dots + a_n \cdot a_n$ can be any complex number, but $a_1 \cdot \overline a_1 + \dots + a_n \cdot \overline a_n$ is always a nonnegative real number. This makes this "conjugated" version of the dot product much more useful.
A: You asked

How can we derive $\vec{a}$ and $\vec{b}$ are orthogonal when $\vec{a}\cdot\vec{b}=0$?

The Wikipedia article Dot product section Complex vectors
states that

The angle between two complex vectors is then given by
$$ \cos \theta ={\frac {\operatorname {Re} (\mathbf {a} \cdot \mathbf {b} )}{\left\|\mathbf {a} \right\|\,\left\|\mathbf {b} \right\|}}. $$

Thus, the correct statement is:
$\vec{a}$ and $\vec{b}$ are orthogonal when $ \operatorname {Re} (\vec{a}\cdot\vec{b}) =0. $
The reason for this is that if you express each
complex number as a two-dimensional vector of its
real and complex part, then a $n$-dimensional
complex vector becomes a $2n$-dimensional real
vector and the complex dot product has a real
part which is the dot product of the real vectors.
A: I think that you are not fully recognizing what the dot product of two complex numbers (or vectors) means. Given two complex numbers, say $z_1$ and $z_2$, then the complex product $z_1z_2^*$, where * denotes the conjugate gives both the scalar and vector products. Specifically,
$$\Re\{z_1z_2^*\}=|z_1| \cdot |z_2| \cos(\zeta)=\frac{1}{2} (z_1z_2^*+z_1^*z_2) \\
\Im\{z_1z_2^*\}=|z_1| \cdot |z_2| \sin(\zeta)=\frac{1}{2} (z_1z_2^*-z_1^*z_2)$$
where $\zeta$ is angle between the two vectors.
This should simplify your determination of how to determine if $a$ and $b$ are orthogonal.
