Finding orthogonal vector with complex coefficients A vector $x \in \mathbb{C}^n$ such that $\langle x,x\rangle = 1$ (with respect to the standard inner product $\langle x,y\rangle = \sum_i x_i \overline{y_i}$) is given.
Show that, if all entries of $x$ satisfy $|x_i|\leq \frac{1}{\sqrt{2}}$, then it is possible to find a vector $y \in \mathbb{C}^n$ such that $\langle x,y\rangle = 0$ and $|y_i| = |x_i|$ for all $i$.
I am not able to solve the above problem. I do not understand how to choose the complex arguments of each coefficient $y_i$ such that the vectors are orthogonal, or why the assumption that $|x_i|\leq \frac{1}{\sqrt{2}}$ is necessary here. Is $\frac{1}{\sqrt{2}}$ the largest value such that it is possible to construct the desired vector $y$? 
 A: To avoid a clash of notations, I will use $k$ as the index and reserve $i$ for the imaginary unit.
The condition $|y_k| = |x_k|$ means $y_k$ is of the form $y_k = e^{i\theta_k}x_k$, so the requirement $\langle y,x \rangle = 0$ boils down to solving the equation
$$
\sum_{k=1}^n e^{i\theta_k}|x_k|^2=0
$$
I feel that there should be more elegant arguments for the existence of a solution for this equation, nevertheless, here are my arguments:
The conditions $\langle x,x\rangle =1$ and $|x_i|\leq \frac{1}{\sqrt{2}}$ implies that $n\geq 2$. When $n=2$, $|x_1|^2 = |x_2|^2=\displaystyle\frac{1}{2}$, so the equation above has the obvious solution $\theta_1 = \theta_2+\pi$. Therefore we can assume that $n\geq 3$.
From $\displaystyle|x_i|^2\leq \frac{1}{2}$ we have
$$
|x_1|^2 \leq \sum_{k=2}^n |x_k|^2
$$
$$
|x_2|^2 \leq |x_1|^2 +\sum_{k=3}^n |x_k|^2
$$
so
$$
|x_2|^2 - \sum_{k=3}^n |x_k|^2 \leq |x_1|^2 \leq \sum_{k=2}^n |x_k|^2
$$
Note that the set
$$
\displaystyle \left\{ \sum_{k=2}^n |x_k|^2e^{i\theta_k} \mid \theta_k\in \mathbb{R}\right\}
$$
consists of vectors of length from $|x_2|^2 - \sum_{k=3}^n |x_k|^2$ varying 
 continuously to $\sum_{k=2}^n |x_k|^2$, so the equation above has a solution.
