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

Question

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.

Answer 1

This is not a complete solution; writing a complete solution turned out to be harder than I thought.

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 $n$.