Orthogonal vectors in complex vector space Consider $u,v\in \mathbb{R}^2$ where $u=(2,1), v=(-1, 2)$. $u$ and $v$ are orthogonal since $u\cdot v=0$. If we put them in $\mathbb{C}$, they should be still orthogonal. However, $ \langle u,v \rangle=(2+i)(-1+2i)\ne 0$. I must misunderstand something here. Could anyone explain this? Thanks in advance.
EDIT: $v=(-1, 2)$ should be converted to  $-1 + 2i$ not $-1-2i$.
 A: When you specify a vector space, you always specify two things: a set and a field. When we take our set to be the complex numbers $ \mathbb{C} $, two possible choices for our field are $ \mathbb{R} $ and $ \mathbb{C} $. So we may think of $ \mathbb{C} $ as a vector space with either real scalars or complex scalars. The choice is important when defining an inner product, since an inner product always maps into whichever field you took your vector space over.
If we choose complex scalars, then one inner product on $ \mathbb{C} $ is $ \langle z, w \rangle = z \overline{w} $. As you rightly point out, $ \langle 2 + i, -1 + 2i \rangle \neq 0 $. So $ 2 + i $ and $ -1 + 2i$ are not orthogonal in this inner product space. That shouldn't be surprising, because as vectors they are linearly dependent; we have $ 2 + i = -i(-1 + 2i) $. We really needed complex scalars for this to be true; no real number $ \alpha $ satisfies $ 2 + i = \alpha (-1 + 2i) $.
If we choose real scalars, then the map $ \langle z, w \rangle = z \overline{w} $ is no longer an inner product on our vector space, since in general $ z \overline{w} $ is not a real number. You can still define an inner product
$$ \langle z, w \rangle = \text{Re}(z) \text{Re}(w) + \text{Im}(z) \text{Im}(w). $$
This is just the dot product on $ \mathbb{R}^2 $ in disguise. With this inner product, you can check that $ \langle 2 + i, -1 + 2i \rangle = 0 $, i.e. $ 2 + i $ and $ -1 + 2i $ are orthogonal.
So even though the set we were working with was the same in both cases, we defined two different vector spaces and two different inner product spaces - the notion of orthogonality in each of them is not the same.
A: It's true that $\mathbb R^2$ and $\mathbb C$ model the same geometric object, but it's not true that the inner product on $\mathbb R^2$ model the same geometric quantity with the product on $\mathbb C$. For one thing, inner product must always be real.
Geometrically, inner product is more about the difference of the angles, i.e. $\langle v, \rangle w = |v||w|\cos(\theta)$ where $\theta$ is the angle between $v$ and $w$. Product of complex numbers is about the addition of the arguments: $r_1e^{i\theta_1}r_2e^{i\theta_2} = r_1r_2e^{i(\theta_1+\theta_2)}$.
To use product of complex numbers to solve the problem, we should use division instead of multiplication, because division $\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}$ tells us the angle between them.
In your case, $\frac{2+i}{-1+2i}=\frac{(2+i)(-1+2i)}{5}=\frac{3i}{5}$. Note that its real part is $0$, hence the argument is $90^\circ$ or $\frac{\pi}{2}$.
A: 
If we put them in $\mathbb C$

"Putting them in  $\mathbb C$" is rather imprecise phrasing. We can define a bijection between $\mathbb R^2$ and $\mathbb C$, but that doesn't mean that they are the same mathematical object, or that the complex number that an element of $\mathbb R^2$ gets sent to is the "same" number. Just because we use a pair of real numbers to represent a complex number doesn't mean that a complex number is a pair of real numbers. Just because the "dot product" in $\mathbb R$ and "complex multiplication" in $\mathbb C$ are referred to as "multiplication" doesn't mean they are the same operation.
Note that if you calculate the real part $(2+i)(−1−2i) = -2-2i^2 = -2-2(-1)=-2+2$ you get zero. If you take two linearly independent vectors in $\mathbb R$, and take the complex numbers represented by those pairs of numbers, the real part of their inner product will be zero. You can think of the imaginary part of the inner product as an "extra" part that shows up in the complex world that can still be nonzero, even when the "real inner product" is zero.
