Discontinuities in the FFT
Your parameters aren't correct for producing a whole number of cycles for each component.

For each $i$ the value of $ \frac{\omega_{i}}{\omega_{s}} N $ has to be a multiple of $ 2 \pi $.

Hope this helps.

Ced



Followup:

Suppose your signal is 

$$ x[n] = A \cos( \alpha n + \phi ) $$

If $ \phi \neq 0 $ there is a possible solution which will give you a strictly real valued DFT bins besides one with a whole number of cycles.  The condition that must be met is: $ \alpha N + 2 \phi $ must be a multiple of $ 2 \pi $.

For a whole number of cycles, your only non-zero bins will be at $ k $ and $ N-k $ where $ k $ is the number of cylces in the frame.  There is no leakage.  The bins will be real valued if $\phi = 0 $ or $\phi = \pi $.

For the alternative above, there will be leakage and it will all be real valued.

Edit:  A simpler way of stating this is that your signal has to be symmetric about the center of your interval.



Response to comment:

Sorry, I didn't see your comment until after I posted the followup.

The mathematical answer lies in equations (23) through (25) in my blog article "DFT Bin Value Formulas for Pure Real Tones".

In order to have strictly real valued bin values, the value of $ U $ must be zero.  For whole numbers of cycles, $ V $ is also zero and eq (25) becomes undefined at the relevant bins and you have to use equation (19).  The alternative solution in my followup makes $ U $ zero, but not $ V $.