Are the only local minima of $\angle(v, Av)$ the eigenvectors? Let $A$ be an invertible $n \times n$ complex matrix. For $v \in \mathbb{CP}^{n-1}$, define 
$$d(v) =  \frac{|\langle A \tilde{v}, \tilde{v} \rangle |^2}{ \langle A \tilde{v}, A \tilde{v} \rangle \langle \tilde{v}, \tilde{v} \rangle}$$
where $\langle \ , \ \rangle$ is the standard  Hermitian inner product and $\tilde{v}$ is any lift of $v$ to $\mathbb{C}^n \setminus \{ 0 \}$
So $d(v) \leq 1$, with equality precisely if $v$ is an eigenvector (by Cauchy-Schwarz).

Are the eigenvectors the only local maxima of $d$?

Motivation: If this is true, than we can prove that complex matrices have complex eigenvectors by a proof analogous to the standard proof that real symmetric matrices have real eigenvectors.
 A: Thank you for this interesting question! (Long time I was expecting the opposite answer.)
True motivation. Let $V$ be a finite-dimensional $\mathbb C$-linear space equipped with a positive-definite hermitian form $\langle-,-\rangle$. Pick a $1$-dimensional $\mathbb C$-linear subspace $p\subset V$. The orthogonal decomposition $V=p\oplus p^\perp$ provides the natural identification and inclusion in
$$\text{T}_p{\mathbb P}_{\mathbb C}V=\text{Lin}_{\mathbb C}(p,V/p)=\text{Lin}_{\mathbb C}(p,p^\perp)\subset\text{Lin}_{\mathbb C}(L,L).$$ The rule $\langle t_1,t_2\rangle:=\text{tr}(t_1\circ t_2^*)$ defines a positive-definite hermitian form on $\text{Lin}_{\mathbb C}(L,L)$ and thus induces a hermitian structure on ${\mathbb P}_{\mathbb C}V$ known as Fubini-Study. The Fubini-Study distance is well known to satisfy the inequalities $0\le\text{dist}(p_1,p_2)\le\frac\pi2$ and the identity
$$\cos\text{dist}(p_1,p_2)=\frac{\langle p_1,p_2\rangle\langle p_2,p_1\rangle}{\langle p_1,p_1\rangle\langle p_2,p_2\rangle}=:\text{ta}(p_1,p_2)$$
(see, for instance, arXiv:0702714).
Question. Are the fixed points of $A$ the only local minima of $\text{dist}(Ap,p)$, where $A$ is an arbitrary holomorphic automorphism of ${\mathbb P}_{\mathbb C}V$ ?
Answer. Suppose that a local maximum $x=p$ of $\text{ta}(Ax,x)$ is not fixed by $A$. Taking suitable representatives $p\in V$ and $A\in\text{GL}_{\mathbb C}V$, we assume
$\langle p,p\rangle=\langle Ap,Ap\rangle=1$ and $g:=\langle Ap,p\rangle\ge0$. If $g=0$, then $\langle Ax,x\rangle=0$ for all $x\in V$ sufficiently close to $p$, hence, for all $x\in V$. Taking an eigenvector $x$ of $A$, we arrive at a contradiction. So, $0<g<1$ (because the hermitian form on $V$ is positive-definite).
Since
$\dim_{\mathbb C}p^\perp+\dim_{\mathbb C}({\mathbb C}p+{\mathbb C}Ap)>\dim_{\mathbb C}V$,
there exists $0\ne w\in p^\perp$ such that
$(1-gA)w\in{\mathbb C}p+{\mathbb C}Ap$. We will show that, for some
$0\ne c\in{\mathbb C}$, an arbitrarily small deformation $p'=p+tcw$ of $p$, $t\in{\mathbb R}$, provides $\text{dist}(Ap',p')<\text{dist}(Ap,p)$. We assume $\langle w,w\rangle=1$ and write $w-gAw=ap+bAp$ for some $a,b\in{\mathbb C}$. It follows that $g\langle Aw,p\rangle+a+bg=0$.
For any $v\in p^\perp$ such that $\langle v,v\rangle=1$, we define
$$w_1(t):=\big\langle A(p+tv),p+tv\big\rangle\big\langle p+tv,A(p+tv)\big\rangle,$$
$$w_2(t):=\big\langle A(p+tv),A(p+tv)\big\rangle(1+t^2)$$
so that
$$\text{ta}\big(A(p+tv),p+tv\big)=\frac{w_1(t)}{w_2(t)}=:\varphi(t).$$
The fact that $x=p$ is a local maximum of $\text{ta}(Ax,x)$ implies $\varphi'(0)=0$ and $\varphi''(0)\le0$.
Taking derivatives with respect to $t$, we obtain
$$w'_1(t)=2\text{Re}\Big(\big(\langle Av,p+tv\rangle+\langle Ap+tAv,v\rangle\big)\langle p+tv,Ap+tAv\rangle\Big),$$
$$w'_2(t)=2\text{Re}\langle Av,Ap+tAv\rangle(1+t^2)+2\langle Ap+tAv,Ap+tAv\rangle t.$$
Therefore,
$$w_1(0)=g^2,\qquad w'_1(0)=2g\text{Re}\big(\langle Av,p\rangle+\langle Ap,v\rangle\big),$$
$$w''_1(0)=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2,$$
$$w_2(0)=1,\qquad w'_2(0)=2\text{Re}\langle Av,Ap\rangle,\qquad w''_2(0)=2\langle Av,Av\rangle+2.$$
The condition $\varphi'(0)=0$ is equivalent to $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$, i.e., to
$2g\text{Re}\big(\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle\big)=0$.
Since it holds for any $v\in p^\perp$, we obtain
$\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle=0$. In particular, $\langle Aw,p\rangle+ag+b=0$, hence,
$g\langle Aw,p\rangle+ag^2+bg=0$. It follows from $g\langle Aw,p\rangle+a+bg=0$ that $a=0$. So, $w-gAw=bAp$ and $\langle Aw,p\rangle=-b$.
The equality $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$ implies $\varphi''(0)=\frac{w''_1(0)w_2(0)-w_1(0)w''_2(0)}{w_2^2(0)}$. As
$$w''_1(0)w_2(0)-w_1(0)w''_2(0)=$$
$$=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2-2g^2\langle Av,Av\rangle-2g^2,$$
we obtain
$2g\text{Re}\langle Av,v\rangle+\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2\le g^2\langle Av,Av\rangle+g^2$,
i.e.,
$$2\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)+\big|\langle Av,p\rangle\big|^2+\big|\langle Ap,v\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$
Replacing $v$ by $uv$ with a suitable unitary $u\in{\mathbb C}$, $|u|=1$, we obtain $\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)\ge0$ and conclude that
$$\big|\langle Av,p\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$
In particular, for $v=w$, we obtain $|b|^2+1-g^2\le|b|^2$. A contradiction.
So, the answer is yes.
