I wouldn't call this the "inverse of a dot product".
Given complex vectors $v = (v_{1}, \ldots, v_{n})$ and $w = (w_{1}, \ldots, w_{n})$, their scalar (=dot) product is given by $v \cdot w = \sum_{j=1}^{n} v_{j} \overline{w}_{j} = v_{1} \overline{w}_{1} + \cdots + v_{n} \overline{w}_{n}$. Why? Well, you want to generalize the usual dot product on $\mathbb{R}^{n}$, but you also want that $v \cdot v \geq 0$ for all $v$, and the vector $(i,0,\ldots,0)$ shows that you can't do without the conjugation. You might ask why, I don't care if $v \cdot v \geq 0$ for all $v$. Most people do: the expression $\|v\| = \sqrt{v \cdot v}$ should define a norm and $\|v - w\|$ a metric on $\mathbb{C}^{n}$, and taking square-roots of non-positve numbers (or even complex numbers) simply isn't well-defined. Note that $\|v\| = \sqrt{v \cdot v} = \sqrt{\sum_{j=1}^{n} |v_{j}|^{2}}$.
Having settled this, let $v$ be a non-zero vector. Its orthogonal complement $U = v^{\perp} = \{u \in \mathbb{C}^{n}\,:u \cdot v = 0\}$ is the set of all vectors orthogonal to $v$. Since $U$ is determined by the single linear equation $u_{1}\overline{v}_{1} + \cdots + u_{n}\overline{v}_{n} = 0$, it is an $(n-1)$-dimensional subspace of $\mathbb{C}^{n}$. Finding solutions is easily achieved using Gauss elimination, this will give you vectors $u_{1}, \cdots, u_{n-1}$, which you can make into an orthonormal basis of $U$ using
Gram-Schmidt (note that the notation $\langle u, v \rangle = u \cdot v$ is to be understood. The fact that you're working with $\mathbb{C}$ and not with $\mathbb{R}$ is immaterial, just be careful to note that $v \cdot (\lambda w) = \overline{\lambda} (v \cdot w)$, i.e., the dot product is conjugate-linear in the second variable.
Finally, in order to solve the equation $w \cdot v = d$ simply take any $u \in U = v^{\perp}$ and put $w = u + \frac{d}{v \cdot v} v$, and note that $w \cdot v = (u + \frac{d}{v \cdot v}v) \cdot v = (u \cdot v) + \frac{d}{v \cdot v} v \cdot v = 0 + d = d$.
Euler's Formula is as follows;
$$e^{i\theta} = \cos\theta + i \sin \theta.$$
You can think of the number $1$ as $e^{\ln1}$ and hence of $1^i$ as $e^{i\ln1}$. But $\ln 1 = 0,$ so really you have $e^{0i} = e^0 = 1$, as I'm sure you're aware. Even if we look at this by feeding $0$ into Euler's Formula, we have
$$e^{0i} = \cos(0) + i\sin(0) = 1.$$
Best Answer
Complex numbers can be used in 2D geometry, but do not generalize to 3D. There is a generalization to 4D, called quaternions.
Vectors can be defined with any number of dimensions, and important cases are 2D and 3D. Algebra on vectors and on complex numbers partially overlap, but not much.