Which complex numbers can we hit with finite sums of solutions to $xy = 1$? Let $z$ be a complex number construed as $x+iy$.
I'm thinking about polynomials of the form $\mathbb{R}[x, y]$ and which parts of the complex plane we can reach by taking finite sums of solutions. Polynomials of this form can also be thought as polynomials in $\mathbb{R}[z, \bar{z}]$ since $x = \frac{1}{2}(z + \bar{z})$ and $y = \frac{1}{2}(z - \bar{z})$.
Anyway, finite sums of solutions to $x^2 + y^2 = 1$ can reach the entire plane.
If our desired complex number $z$ is zero, then we're done.
If our desired complex number $z$ has magnitude $|z|$, then we can walk to $|z|$ if $|z|$ is an integer or walk to $\lfloor \; |z| \; \rfloor$, jump up and then jump down in such a way that we cover $|z| - \lfloor \; |z| \; \rfloor$ horizontal distance.

0 - - - - - / \ dest

Then we multiply that entire sum by $z/|z|$ componentwise, which gives us a finite sum of complex numbers of magnitude one.
For the polynomial equation $y = 0$, we can only reach the line $y = 0$ by taking finite sums.
For the polynomial equation $xy = 0$, we can reach anywhere in $\mathbb{C}$ in at most two steps by jumping horizontally and then jumping vertically.
I'm curious where all we can reach by considering finite sums of solutions to $xy = 1$, since the size of the steps we take becomes huge when our direction approaches horizontal or vertical.
 A: As you consider the constraint $xy=1$, the complex numbers at our disposal have the form $z = x+iy = x + \frac{i}{x}$. Now we would like to construct complex numbers $\zeta = a+ib$ from finite sums of the previous ones.
Given that $z$ possesses one parameter $x$, when $\zeta$ has two ($a$ and $b$), let's try what kind of complex numbers can be constructed with the help of two $z$-numbers only :
$$
\begin{array}{rcl}
\zeta = z_1+z_2 
   &\Leftrightarrow&
   a+ib = x_1+x_2+i\left(\frac{1}{x_1}+\frac{1}{x_2}\right) \\
   &\Leftrightarrow&
   \begin{cases}
      a = x_1+x_2 \\
      b = \displaystyle\frac{1}{x_1}+\frac{1}{x_2}
   \end{cases}
   \verb+ +\\
   &\Leftrightarrow&
   \begin{cases}
      x_1 = a-x_1 \\
      0 = bx_2^2-abx_2-a
   \end{cases}
   \verb+ +\\
   &\Leftrightarrow&
   \begin{cases}
      x_1 = a-x_2 \\
      x_2 = \displaystyle\frac{ab\pm\sqrt{a^2b^2-4ab}}{2b}
   \end{cases}
\end{array}
$$
As $x_2$ should be real, there is no solution when $\Delta := a^2b^2-4ab = (ab-2)^2-4 < 0$; in other words, two $z$-numbers are sufficient to generate any $\zeta = a+ib$ with $(ab-2)^2-4 \ge 0$.
Let's see if adding a third number permits to lift this constraint :
$$
\begin{array}{rcl}
\zeta = z_1+z_2+z_3 
   &\Leftrightarrow&
   a+ib = x_1+x_2+x_3+i\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right) \\
   &\Leftrightarrow&
   \begin{cases}
      a = x_1+x_2+x_3 \\
      b = \displaystyle\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}
   \end{cases}
   \verb+ +\\
   &\Leftrightarrow&
   \begin{cases}
      x_1 = t \\
      x_2 = a-t-x_3 \\
      0 = (1-bt)x_3^2-(1-bt)(a-t)x_3-t(a-t)
   \end{cases}
\end{array}
$$
with $t$ a free parameter. Then, in the same way as before, $x_3$ is real only in the cases where $\Delta := (1-bt)^2+4(a-t) \ge 0$; this condition can be always satisfied, since $t$ is arbitrary.
In conclusion, you need at most three terms in your sum and spend a Happy New Year !
A: Here is a proof assisted by a graphical representation.
We are interested in numbers of the form $x+\frac{i}{x}$ (that we will call $h$-terms, letter $h$ standing for "hyperbolic") and their finite sums. Having seen the excellent analytical answer of Abezhiko, I wondered how to represent complex numbers of the form of the sum of only 2 $h$-numbers:
$$z=x+iy=\left(a+\frac{i}{a}\right)+\left(b+\frac{i}{b}\right) \ \iff \ \begin{cases}x&=&a+b\\ y&=&\frac{1}{a}+\frac{1}{b}\end{cases}$$
Otherwise said:
$$\begin{cases}b&=&x-a\\ y&=&\frac{1}{b}+\frac{1}{a}\end{cases}$$
which means that points $(x,y)$ are situated on hyperbolas with common equation :
$$y=\frac{1}{x-a}+\frac{1}{a}=f_a(x)\tag{1}$$
Please note that $f_a(0)=0$ for any $a$ ; as a consequence, all these hyperbolas pass through the origin.
Allowing $a$ to take all possible values $\ne 0$, almost all the plane is "swiped" by these hyperbolas that we can recognize in the green regions of the following representation :

Fig. 1: The four green regions (occupied by the sums of 2 $h$-terms) are defined by $xy>4$ (quadrants I and III) and $xy<0$ (quadrants II and IV).
The excluded zone is the interior ($0 < xy < 4$) of a hyperbola defined by $y=\frac{4}{x}$ (see the proof in Appendix 2 below).
From this result, it is not difficult to establish that the addition of a third $h$-term will allow the "occupation" of the whole plane (see Appendix 1) (the $y$ axis deserving a special treatment).
Appendix 1: (the case where we need to add a third $h$-term) Let us consider the case of a point $u+iv$ on the white upper right area of the figure, i.e., with $u,v>0$ and $uv<4$. Let us look if we can add to it an $h$-term $x+\frac{i}{x}$ such that
$$(u+iv)+(x+\frac{i}{x})=(u+x)+i(v+\frac{1}{x}) \ \ \text{belongs to the green region in Quad. I}\tag{2}$$
Otherwise said, is it possible to find $x>0$ such that :
$$(u+x)(v+\frac{1}{x})>4 \ \iff \ vx^2+(uv-3)x+u>0$$
which is verified for any large enough $x$ because $v>0$.
Conclusion: (2) means that there exists $x$ such that
$$(u+iv)+(x+\frac{i}{x})=(a+\frac{i}{a})+(b+\frac{i}{b})$$
finally giving $u+iv$ as an (algebraic) sum of three $h$-terms.
Appendix 2: Why is $y=\frac{4}{x}$ the envelope ? A classical result says that the envelope of a family of curves with equation $y=f_a(x)$ is obtained by eliminating parameter $a$ in the system obtained by coupling the initial parametric equation and its derivative with respect to the parameter (considering $x$ and $y$ as constants) :
$$\begin{cases}y&=&f_a(x)\\0&=&\frac{\partial f_a}{\partial a} \end{cases} \ \iff \ \begin{cases}y&=&\frac{1}{x-a}+\frac{1}{a}\\0&=&\frac{1}{(x-a)^2}-\frac{1}{a^2} \end{cases}$$
The second equation gives
$$x-a=\pm a\tag{3}$$

*

*In the case of plus sign in (3), one gets $a=\frac{x}{2}$ which, inserted into the first equation, gives :

$$y=\frac{1}{x-\tfrac{x}{2}}+\frac{2}{x}=\frac{4}{x}$$
as desired.

*

*In the case of a minus sign in (3), one gets $x=0$ and $y=0$, which are the equations of the two axes, clearly parts of the envelope too!

