Consider $z\in\mathbb{C}$ to be a non-zero algebraic number such that $\operatorname{Im} z = \operatorname{Re} z$ or $\operatorname{Im} z = -\operatorname{Re} z$.
The question is simple: Is the degree of this number (over $\mathbb{Q}$) guaranteed to be even?
My gut feeling is that yes, it has to be even, but I can't figure out a straightforward way to see this.
Yes. This is true. Let $L=\Bbb{Q}(z)$ and let $K=L\cap\Bbb{R}$ be the real subfield. Obviously $z\notin K$, so $[L:K]>1$.
Let $a=z^4\in K$. Consider the polynomial $f(x):=x^4-a\in K[x]$ that has $z$ as a root. If $f(x)$ is irreducible over $K$, then $[L:\Bbb{Q}]$ is divisible by $[K(z):K]=4$ and we are done. We need to study the case that $f(x)$ isn't irreducible over $K$ in detail.
By a result that deserves to be better known, locally explained here, this means that either $a$ is a square of an element of $K$, say $a=b^2$, or $a$ is of the form $a=-4c^4$ for some element $c\in K$.
It follows in either case that $z$ is a zero of a quadratic in $K[x]$. In the former case $$ x^4-a=(x^4-b^2)=(x^2-b)(x^2+b) $$ and in the latter case $$ x^4-a=(x^4+4c^4)=(x^2-2cx+2c^2)(x^2+2cx+2c^2). $$ We have shown that $[K(z):K]>1$ as well as $[K(z):K]\le2$, so when $f(x)$ is reducible we can conclude that $[L:\Bbb{Q}]$ is a multiple of $[K(z):K]=2$ again proving the claim
Actually we do not require the full power of the linked theorem to conclude. The roots of $f(x)$ have the form $zi^k, k=0,1,2,3$. None of them are real. Thus the only factors of $f(x)$ in $\Bbb{R}[x]$ are $(x-z)(x-\overline{z})$ and the other factor with complementary zeros. So should $f(x)$ be reducible over $K$, we have shown that the minimal polynomial of $z$ over $K$ is then quadratic.
In the same vein, the former case (of the first proof) $a=b^2, b\in K$, cannot occur because $z^2$ is pure imaginary. The Sophie Germain style (or Aurifeuillian) factorization in the case $a=-4c^4$, on the other hand, is a live possibility.
Let $w$ be a nonzero algebraic number such that $\operatorname{Im} w = \pm\operatorname{Re} w$.
Clearly $w\not\in\mathbb{Q}$.
Suppose $[\mathbb{Q}(w):\mathbb{Q}]$ is odd.
Since $[\mathbb{Q}(w):\mathbb{Q}]$ is odd, it follows that $i\not\in\mathbb{Q}(w)$, so $[\mathbb{Q}(w,i):\mathbb{Q}(w)]=2$.
Then from the chain $$ \mathbb{Q} \subset \mathbb{Q}(w) \subset \mathbb{Q}(w,i) $$ it follows that $[\mathbb{Q}(w,i):\mathbb{Q}]$ is even but not a multiple of $4$.
Let $a=\operatorname{Re} w$.
By hypothesis, we have $w=a(1\pm i)$.
Then $a$ is a nonzero real algebraic number, and $\mathbb{Q}(a,i)=\mathbb{Q}(w,i)$.
Since $a$ is real, we have $[\mathbb{Q}(a,i):\mathbb{Q}(a)]=2$.
Then from the chain $$ \mathbb{Q} \subseteq \mathbb{Q}(a^4) \subseteq \mathbb{Q}(a^2) \subseteq \mathbb{Q}(a) \subset Q(a,i) = \mathbb{Q}(w,i) $$ it follows that $$ \mathbb{Q}(a^4) = \mathbb{Q}(a^2) = \mathbb{Q}(a) $$ else $[\mathbb{Q}(w,i):\mathbb{Q}]$ would be a multiple of $4$.
But from the identity $$ w^4 = -4a^4 $$ we get $a^4\in \mathbb{Q}(w)$, so $\mathbb{Q}(a^4)\subset\mathbb{Q}(w)$.
Then from $\mathbb{Q}(a)=\mathbb{Q}(a^4)$, we get $a\in\mathbb{Q}(w)$, hence from $$ w=a(1\pm i) $$ we get $i\in\mathbb{Q}(w)$, contradiction.
Hence $[\mathbb{Q}(w):\mathbb{Q}]$ must be even.
