If $F=\Bbb C$, prove that there exists a unit vector $u$ in $V$ and a complex number $\theta$ with $|\theta|=1$ such that $H_u(x)=\theta y$ 
Let $V$ be a finite-dimensional inner product space over $F$. Let $x$ and $y$ be linearly independent vectors in V such that $\|x\|=\|y\|$. If $F=\Bbb C$, prove that there exists a unit vector $u$ in $V$ and a complex number $\theta$ with $|\theta|=1$ such that $H_u(x)=\theta y$, where $H_u(x)=x-2\langle x,u\rangle u$.

There is a hint: Choose $\theta=\langle x,y\rangle$, but how to prove that $|\theta|=1?$
If I select $u=\frac{x-\theta y}{\|x-\theta y\|}$, then
$$H_u(x)=x-\frac{2\|x\|^2-2\langle x,\theta y\rangle}{\|x-\theta y\|}(x-\theta y)=\\x-\frac{\|x\|^2+ \|\theta y\|^2-2\langle x,y\rangle}{\|x-\theta y\|}(x-\theta y)$$
How to get the second equality?
Thanks for helping.
 A: Choose $\theta=\frac{\langle x,y\rangle}{\lvert\langle x,y\rangle\rvert}$.
$$\lvert\theta\rvert= \frac{\lvert\langle x,y\rangle\rvert}{\lvert\langle x,y\rangle\rvert}=1$$
$$\langle x,\theta y\rangle= \overline{\theta}\langle x,y\rangle=\frac{\langle y,x\rangle}{\lvert\langle x,y\rangle \rvert}\langle x,y\rangle=\frac{\lvert\langle x,y\rangle\rvert^2}{\lvert\langle x,y\rangle\rvert}=\lvert\langle x,y\rangle\rvert$$
So we have that $\langle x,\theta y\rangle$ is a real number. Therefore $\overline{\langle x,\theta y\rangle}=\langle x,\theta y\rangle$
$$\|x-\theta y\|^2=\langle x-\theta y,x-\theta y\rangle=
\langle x,x\rangle-\langle x,\theta y\rangle-\langle\theta y,x\rangle+\langle\theta y,\theta y\rangle=
\|x\|^2-\langle x,\theta y\rangle-\overline{\langle x,\theta y\rangle}-\lvert\theta\rvert\|y\|^2=\|x\|^2-\langle x,\theta y\rangle-\langle x,\theta y\rangle+\|y\|^2=\|x\|^2-2\langle x,\theta y\rangle+\|x\|^2=2\|x\|^2-2\langle x,\theta y\rangle$$
By choosing $u=\frac{x-\theta y}{\|x-\theta y\|}$
$$H_u(x)=x-2\langle x,u\rangle u= x-2\langle x,\frac{x-\theta y}{\|x-\theta y\|}\rangle\frac{x-\theta y}{\|x-\theta y\|}=
x-\frac{2}{\|x-\theta y||^2}(\langle x,x\rangle-\langle x,\theta y\rangle)(x-\theta y)$$
$$=x-\frac{2}{2\|x\|^2-2\langle x,\theta y\rangle}(\|x\|^2-\langle x,\theta y\rangle)(x-\theta y)= x-1(x-\theta y) = x-x+\theta y=\theta y$$
Giving us our desired answer.
