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.
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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.
Best Answer
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 !