# Complex numbers manipulation

I have a complex number s of the form s=[1/sqrt(NN$_{t})] e^{j\phi_{k}}$ S$_{o}$ is another complex number. It is given that |s-s$_{o}$|<= $\epsilon$ where 0< $\epsilon$< 2

It is stated in the literature that this constrain can be written as $$\phi_{k}=\arg s \in\left[\gamma , \gamma + \delta \right]$$ where $\gamma$ and $\delta$ are given by $\gamma$ =arg S$_{o}$ - arccos(1- $\epsilon ^2/2)$ and $\delta$ =2arccos(1- $\epsilon^2/2)$ Can anyone explain how is that possible?

• I adjusted your display equation to try to make it say what I think you wanted to say. If I got it wrong, feel free to revert my edits. – The Photon Oct 8 '14 at 19:22

This is only possible if $|s|=|s_0|=1$. Squaring the original inequality gives

$$|s-s_0|^2=|s|^2+|s_0|^2-2\cos(\Delta\phi)\tag{1}\le\epsilon^2$$

where $\Delta\phi=\arg\{s\}-\arg\{s_0\}$ is the phase difference between $s$ and $s_0$. If $|s|=|s_0|=1$ is satisfied we get from (1)

$$2(1-\cos(\Delta\phi))\le\epsilon^2$$

or, equivalently,

$$\cos(\Delta\phi)\ge1-\frac{\epsilon^2}{2}\tag{2}$$

From this inequality it follows that

$$|\Delta\phi|\le\arccos\left(1-\epsilon^2/2\right)$$

which is equivalent to the condition in your question.

• I got the lower limit for $\phi _{k}$ from your equation but I couldn't get upper limit – Aashish Sharma Oct 9 '14 at 4:22
• The upper limit is just $\arg\{s_0\}+\arccos(1-\epsilon^2)$, and the lower limit is $\arg\{s_0\}-\arccos(1-\epsilon^2)$, so the difference between the two arguments is never greater than $\arccos(1-\epsilon^2)$. – Matt L. Oct 9 '14 at 7:18
• Does it mean they have assumed maximum magnitude of arg{S $_{o}$} is arccos(1- $\epsilon ^2/2)$ – Aashish Sharma Oct 9 '14 at 9:07