# Is the real-imaginary Schanuel's conjecture equivalent to the full Schanuel's conjecture?

Schanuel's conjecture says the following about the transcendence of numbers related by the complex exponential function:

Given any $$n$$ complex numbers $$z_1, ... z_n$$ that are linearly independent over the rational numbers $$\mathbb{Q}$$, the field extension $$\mathbb{Q}(z_1, ..., z_n, e^{z_1}, ..., e^{z_n})$$ has transcendence degree at least $$n$$ over $$\mathbb{Q}$$.

I would like to control the real and imaginary parts of the numbers separately. Thus I'm interested in the restriction to real and imaginary numbers $$z_1, ... z_n$$. Since for imaginary numbers $$yi$$, the function values $$e^{yi}$$, $$\cos y$$ and $$\sin y$$ are definable from each other by algebraic functions, the real-imaginary Schanuel's conjecture can be stated using only real functions as follows:

Given any $$m+n$$ real numbers $$x_1, ... x_m, y_1, ... y_n$$ such that each of the sets $$x_1, ... x_m$$ and $$y_1, ... y_n$$ are linearly independent over the rational numbers $$\mathbb{Q}$$, the field extension $$\mathbb{Q}(x_1, ... x_m, y_1, ... y_n, e^{x_1}, ..., e^{x_m}, \cos y_n, ... \cos y_n)$$ has transcendence degree at least $$n$$ over $$\mathbb{Q}$$.

Is this conjecture equivalent to the full Schanuel's conjecture? I know that it's implied by Schanuel's conjecture since $$x_1, ... x_m, y_1i, ... y_ni$$ are linearly dependent, that is the equation $$a_1x_1+ ... +a_mx_m+b_1i+ ... +b_ny_ni=0$$ has a rational solution $$a_1, ..., a_m, b_1, ..., b_n$$ iff it is a simultaneous solution of the two equations $$a_1x_1+ ... +a_mx_m=0$$ and $$b_1+ ... +b_ny_n=0$$. In the converse direction we have $$\overline{\mathbb{Q}(z_1, \overline{z_1}, ..., z_n, \overline{z_n}, e^{z_1}, e^{\overline{z_1}}, ..., e^{z_n}, e^{\overline{z_n}})}=\overline{\mathbb{Q}(\Re z_1 \Im z_1, ..., \Re z_n, \Im z_n, e^{\Re z_1}, e^{\Im z_1}, ..., e^{\Re z_n}, e^{\Im z_n})}$$ (where $$\overline{F}$$ denotes the algebraic closure of the field $$F$$, $$\overline{z}$$ denotes the complex conjugate of $$z$$, $$\Re z$$ denotes the real part of $$z$$, and $$\Im z$$ denotes the imaginary part of $$z$$), but I'm having trouble proving that linear independence of $$z_1, \overline{z_1}, ..., z_n, \overline{z_n}$$ is equivalent to linear independence of $$\Re z_1 \Im z_1, ..., \Re z_n, \Im z_n$$. Additionally, I'm not sure if I can prove that Schanuel's conjecture for $$n$$-tuples closed under complex conjugation is equivalent to Schanuel's conjecture for all $$n$$-tuples.

I'm having trouble proving that linear independence of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ is equivalent to linear independence of $$\Re z_1, \Im z_1, ..., \Re z_n, \Im z_n$$.
Assume as induction hypothesis that $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ and $$\Re z_1, \Im z_1, ..., \Re z_n, \Im z_n$$ span the same vector subspace of $$\mathbb{C}$$ over $$\mathbb{Q}$$. Then there are four possibilities for whether the real and imaginary parts of $$z_{n+1}$$ is linearly independent with any maximal linearly independent subset of $$z_1, ..., z_n$$:
1. $$\Re z_{n+1}$$ and $$i\Im z_{n+1}$$ are both linear combinations of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$. By the induction hypothesis it is equivalent that they are both linear combinations of $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, meaning that $$\Re z_{n+1}$$ is a linear combination of $$\Re z_1, ..., \Re z_n$$ and $$\Im z_{n+1}$$ is a linear combination of $$\Im z_1, ..., \Im z_n$$. Thus $$\Re z_{n+1}$$ and $$\Im z_{n+1}$$ are both in the vector space spanned by $$\Re z_1, \Im z_1, ..., \Re z_n, \Im z_n$$.
2. $$\Re z_{n+1}$$ is linearly independent with $$\Re z_1, ..., \Re z_n$$, thus linearly independent with $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus linearly independent with $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$, whereas $$\Im z_{n+1}$$ $$\Im_{n+1}$$ is a linear combination of $$\Im z_1, ..., \Im z_n$$, thus $$i\Im_{n+1}$$ is a linear combination of $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus $$i\Im_{n+1}$$ is a linear combination of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$. Then $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}$$, $$z_1,\overline{z_1},...,z_n,\overline{z_n}, \overline{z_{n+1}}$$, $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}, \overline{z_{n+1}}$$, and $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n, \Re z_{n+1}$$ span the same vector space, whose dimension is $$1$$ greater than the dimension of the vector space spanned by either of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ or $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$.
3. $$\Re z_{n+1}$$ is a linear combination of $$\Re z_1, ..., \Re z_n$$, thus a linear combination of $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus a linear combination of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$, whereas $$\Im z_{n+1}$$ $$\Im_{n+1}$$ is linearly independent with $$\Im z_1, ..., \Im z_n$$, thus $$i\Im_{n+1}$$ is linearly independent with $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus $$i\Im_{n+1}$$ is linearly independent with $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$. Then $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}$$, $$z_1,\overline{z_1},...,z_n,\overline{z_n}, \overline{z_{n+1}}$$, $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}, \overline{z_{n+1}}$$, and $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n, \Im z_{n+1}$$ span the same vector space, whose dimension is $$1$$ greater than the dimension of the vector space spanned by either of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ or $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$.
4. $$\Re z_{n+1}$$ is linearly independent with $$\Re z_1, ..., \Re z_n$$, thus linearly independent with $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus linearly independent with $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$, and $$\Im z_{n+1}$$ $$\Im_{n+1}$$ is linearly independent with $$\Im z_1, ..., \Im z_n$$, thus $$i\Im_{n+1}$$ is linearly independent with $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$, thus $$i\Im_{n+1}$$ is linearly independent with $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$. By the induction hypothesis $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ and $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$ span the same vector space, and $$a_1 z_1, ..., a_n z_n, a_{n+1} z_{n+1} =0$$ is equivalent to $$a_1 \Re z_1, ..., a_n \Re z_n, a_{n+1} \Re z_{n+1} =0$$ and $$a_1 \Im z_1, ..., a_n \Im z_n, a_{n+1} \Im z_{n+1}$$ simultaneously, so the dimension of the vector space spanned by either of $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}, \overline{z_{n+1}}$$, and $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n, \Im z_{n+1}$$ is $$2$$ greater than the dimension of the vector space spanned by either of $$z_1,\overline{z_1},...,z_n,\overline{z_n}$$ or $$\Re z_1, i\Im z_1, ..., \Re z_n, i\Im z_n$$.
In all cases, $$z_1,\overline{z_1},...,z_n,\overline{z_n}, z_{n+1}, \overline{z_{n+1}}$$ and $$\Re z_1, \Im z_1, ..., \Re z_n, \Im z_n, \Re z_{n+1}, \Im z_{n+1}$$ span the same vector subspace. By induction (the base case is trivial) this applies to every $$n$$-tuple of complex numbers for every natural number $$n$$.