I think the best way to understand this is via projective geometry. The idea is that in quantum mechanics we always assume that our state vector is normalized and we don't care about global phases, so essentially our states are not defined by vectors, but rather by mathematical objects called rays. Rays are defined by starting out with all nonzero vectors of some fixed vector space, and then identifying all vectors that are scalar multiples of one another. The space of all states is known as projective space.
More concretely, $n-1$-dimensional complex projective space is defined as follows:
$$ \mathbb{CP}^{n-1} = \left( \mathbb{C}^n \setminus \left\{ 0 \right\} \right) / \sim $$
where $ \sim $ is the equivalence relation on $ \mathbb{C}^n \setminus \left\{ 0 \right\} $ that identifies two nonzero vectors $ u,v $ iff there exists some nonzero scalar $ \lambda \in \mathbb{C}^* $ such that $ u = \lambda v $. The simplest nontrivial example of a complex projective space is given by $n=2$, which is precisely what we need for a qubit. The resulting space, $ \mathbb{CP}^1 $, is topologically a sphere, which can be identified with the Riemann sphere - i.e. the complex plane with an additional point at infinity. To see this, we write the points of $ \mathbb{CP}^1 $ as:
$$ \left[ a : b \right] , \quad a,b \in \mathbb{C}, \; \left( a, b\ \right) \neq \left( 0, 0 \right) , $$
where the notation helps us remember that $ \left[ a : b \right] $ should be thought of representing an equivalence class, in the sense that $ \left[ a : b \right] = \left[ \lambda a : \lambda b \right] $ for any nonzero $ \lambda $.
For example, now we can see that:
$$ \left[ 1/\sqrt{2} : -1/\sqrt{2} \right] = \left[ -i/\sqrt{2} : i/\sqrt{2} \right] $$
with $ \lambda = -i $. This is the mathematical context in which the two vectors you obtained are indeed equal: they are not equal as vectors, but they represent the same ray, so they correspond to the same point of the projective space $ \mathbb{CP}^1 $.
Keep in mind that a point of $ \mathbb{CP}^1 $ represents a whole ray (line) of $ \mathbb{C}^2 $; somewhat confusingly, $ \mathbb{CP}^1 $ is often referred to as the complex projective line (since it is one-dimensional as a complex manifold / variety).
Using this language, we note that if $ a \neq 0 $ we can always make it $1$ (multiplying both coordinates by $ \lambda = a^{-1} $); so we get a subspace of $ \mathbb{CP}^1 $ defined by fixing $ a = 1 $. Assuming this "normalization", $b$ can obtain any complex value (including zero), and for any two distinct values $ b_1 \neq b_2 $ we have $ \left[ 1 : b_1 \right] \neq \left[ 1 : b_2 \right] $. So this subspace where $ a=1 $ is just a copy of $ \mathbb{C} $.
For $ a = 0$, we only obtain a single point, since in that case $ b $ is not allowed to be zero, and we have $ \left[ 0 : b_1 \right] = \left[ 0 : b_2 \right] $ for any two nonzero $ b_1, b_2 $. Thus, we see that $ \mathbb{CP}^1 $ is obtained from the complex plane by adding a single "point at infinity". This added point "closes" the plane on all directions, thus forming a sphere (this process is known as one-point compactification). For example, we can think of the point of infinity as being located at the south pole (an arbitrary choice, of course). There are several ways to see the correspondence with a sphere more concretely. I'll show two of those - one purely mathematical, and the other more "physical".
- Stereographic projection: start with a vector:
$$ \begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb{C}^2 , $$
or equivalently - start with the ray represented by $ \left[ a : b \right] $.
If $ a = 0$ then this vector maps to the south pole, so we are done. Otherwise, multiply the vector by $ \lambda = a^{-1} $, to find that $ \left[ a : b \right] = \left[ 1 : a^{-1}b \right] $, and denote $ z := a^{-1}b $. If $ z=0 $ we are done again, since we got the north pole. Otherwise, $z$ can be written uniquely as $ z = r e^{i \phi} $ with $ r>0, 0 \leq \phi <2 \pi $. This $ \phi $ is our azimuthal angle, and the zenith angle $ \theta $ is obtained by:
$$ \theta = 2\cdot \arctan (r) .$$
Note that this strange function maps the inside of the unit circle $ \left\vert z \right\vert^2 < 1 $ to the northern hemisphere, the unit circle itself to the equator, and the rest of the entire complex plane to the southern hemisphere.
Let's see what we get in your example: starting with $ \left[ -\frac{i}{\sqrt{2}} : \frac{i}{\sqrt{2}} \right] $, we multiply by $ \lambda = \sqrt{2}i $ to get $ \left[ 1 : -1 \right] $. So, $ z = -1 = 1 \cdot e^{i\pi} $. $ \arctan (1) = \pi/4 $, so after multiplying by $2$ we can see that we are on the equator ($\theta = \pi/2$) with $ \phi=\pi $. In Cartesian coordinates, that would be:
$$ x = \sin \theta \cos \phi = -1, \quad y=z=0 .$$
But how can we tell if we chose the "correct" mapping? What does the position on the sphere really mean physically? That leads us to -
- The Bloch vector as quantum expectation values: starting with any normalized vector $ \vert \psi \rangle \in \mathbb{C}^2 $ (I switched to bra-ket notation since we'll now be doing physics), we can define the following expectation values:
$$ x := \langle \psi \vert \hat{X} \vert \psi \rangle, \; y := \langle \psi \vert \hat{Y} \vert \psi \rangle, \; z := \langle \psi \vert \hat{Z} \vert \psi \rangle , $$
where $ \hat{X}, \hat{Y}, \hat{Z} $ are the Pauli matrices (perhaps up to normalization). The vector $ \mathbb{r} := \left( x, y, z \right) $ is the Bloch vector, and it turns out that $ \left\vert \mathbb{r} \right\vert^2 = 1 $ (more generally - for a mixed state, $ \left\vert \mathbb{r} \right\vert^2 \leq 1 $). So we think of $\mathbb{r}$ as representing position on the Bloch sphere; note that $\mathbb{r}$ is independent of the global phase, so we did not need to multiply $ \vert \psi \rangle $ by a scalar to obtain a "clean" mapping to a sphere (however, this procedure did depend on $ \vert \psi \rangle $ being normalized).
The neat thing is that the two maps outlined above are in fact the same map. I don't know of any proof that does not require at least some amount of computation, so I will not prove it here. But it is true, and fairly fundamental - in fact, you already have the tools to prove it yourself, which would make good exercise.
All this and much more is explained (with helpful pictures!) in the book "Geometry of Quantum States" by Bengtsson and Życzkowski.
