# Orthogonal complex vectors with same magnitude of each entry

Given a complex vector $z = (z_1, \ldots, z_n)$, how to find a complex vector $y = (y_1, \ldots, y_n)$ such that $z \cdot y = 0$ (with respect to the complex inner product) and $|y_i| = |z_i|$ for each $i$? Is this always possible? If not, then what conditions does $z$ have to satisfy for this to be possible?

• Are you in $\mathbb R$ or in $\mathbb C$? In the first case, this is only possible if there exists a choice of $\pm$ such that $\pm z_1^2\pm z_2^2\pm\ldots\pm z_n^2 = 0$. – amsmath Oct 9 '17 at 18:59
• @amsmath All of the entries are complex in general. – trey Oct 9 '17 at 20:06

It is not always possible to find $y$ such that $z \cdot y = 0$. For example, if $z=(z_1, 0, \cdots, 0)$, where $|z_1| \neq 0$ then $y = (|z_1|e^{i \theta},0,\cdots,0)$, which implies $$z.y = |z_1|^2 e^{i \phi} \neq 0.$$
• An even simpler approach is to take $n=1$, which is essentially what you did. – Zach Boyd Oct 9 '17 at 19:30
The dot product is zero when the 'vectors' are normal. Hence, choose $y=ze^{\pm i\pi/2}=\pm iz$.