= Solution
A <quadratic form> over $\mathbb R$ or $\mathbb C$ is a function $q(x)=B(x,x)$ for a symmetric <bilinear form> $B$, equivalently a homogeneous polynomial of degree two. Its polarization is $B(x,y)=[q(x+y)-q(x)-q(y)]/2$. If $q$ is not identically zero, choose $v$ with $q(v)\ne0$. Every $x$ decomposes as
$$
x=\frac{B(x,v)}{q(v)}v+\left(x-\frac{B(x,v)}{q(v)}v\right),
$$
where the second term lies in $v^\perp$ and $kv\cap v^\perp=\{0\}$. The <quadratic form> is consequently a direct sum of its one-dimensional value on $v$ and its restriction to $v^\perp$. Induction on dimension diagonalizes it. If $q=0$, any basis suffices. Rescale every nonzero diagonal coefficient: over the reals it becomes $1$ or $-1$, while over the complexes a square root makes it $1$. Hence
$$
\boxed{q(x)=\sum_{j=1}^na_jx_j^2,\qquad a_j\in\{-1,0,1\}.}
$$
Over $\mathbb C$ the negative coefficients can all be avoided.
For a real <quadratic form>, let $p$ and $m$ be its positive and negative indices in a diagonalization. Its <rank of a quadratic form> is $p+m$ and its <signature of a quadratic form> is $p-m$, or the ordered pair $(p,m)$ in the alternative convention. These indices are intrinsic: the largest dimension of a subspace where the form is positive definite is $p$, since any larger subspace intersects the nonpositive coordinate subspace nontrivially; similarly for $m$. For the form here, completing squares gives
$$
q=(x_2+x_1+x_3)^2-(2x_1+x_3)^2.
$$
The two displayed linear coordinates, together with $x_1$, form an invertible coordinate change. Therefore
$$
\boxed{\operatorname{rank}q=2,\qquad(p,m)=(1,1),\qquad\operatorname{signature}q=0.}
$$
There is one zero coordinate direction.
The PDF's final dimension condition is $n\ge2^d$, not the fraction produced by the TeX conversion. We prove the requested <common zero of complex quadratic forms> by induction on $d$. A complex <quadratic form> of rank $r$ has canonical expression $x_1^2+\cdots+x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+ie_{2j}$ and add the $n-r$ radical coordinate vectors. Their span is a <totally isotropic subspace>, because the form is zero on every linear combination. This proves that <complex quadratic forms have a large isotropic subspace>: its dimension is
$$
\left\lfloor\frac r2\right\rfloor+n-r=n-\left\lceil\frac r2\right\rceil\ge\left\lfloor\frac n2\right\rfloor.
$$
For $d=0$ there is a nonzero vector whenever $n\ge1$. For $d\ge1$, take such a subspace $E$ for $q_1$. If $n\ge2^d$, then $\dim E\ge2^{d-1}$. Apply the induction hypothesis to the restrictions of $q_2,\ldots,q_d$ to $E$. It supplies a nonzero $x\in E$ annihilated by those forms, and $q_1(x)=0$ by construction. Thus
$$
\boxed{\exists x\ne0:\ q_1(x)=\cdots=q_d(x)=0\quad\text{when }n\ge2^d.}
$$
The argument makes no nondegeneracy assumption on any of the forms.
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