Hello Aisha,
You are looking at a polynomial function of z, with two roots as you stated. To see what is going on, it helps to look first at just the function f(z)=z and its single root at z=0. The upper left figure below shows the phase angle of z, which by convention is -pi<angle<pi. The angle is 0 along the pos real axis, runs ccw to +pi on the neg real axis and cw to -pi, also on the neg real axis. There is a 2pi discontinuity on neg real axis. That is the standard convention, but it's just a convention. You could have 0<angle<2pi if you wanted. Then the angle starts at 0 on the pos real axis and goes all the way around ccw to +2pi on pos real axis. Consequently there is a 2pi discontinuity on the pos real axis.
Once you take a cosine or sine of the phase angle, the discontinuity disappears because trig functions don't care about 2pi discontinuities. So you are not 'solving' the angle discontinuity, it's always there. You are merely inserting the angle into a function that doesn't care about it [as long as the discontinuity is an integral multiple of 2pi]. That's shown on the figure on the upper right, where yellow colors are pos cos and blue colors are neg cos as you mentioned. Note that as you circle the origin, you go from the top to the bottom of the colorbar and back again. I'll call that one circuit of the colorbar.
At z=0, the root of the function, all colors converge and the angle is undefined. That's like taking atan(y/x) where both x and y are zero.
The lower two figures are the function you have. There two roots, so two points where the angle is undefined. The phase angles associated with each point are additive. That means that on the lower right plot, if you take a large radius and go around both roots, you go through two circuits of the colorbar. Picking a smaller path that encircles one of the roots gives one circuit of the colorbar. The same kind of argument applies to lower left plot, where you go through either one or two angle discontinuities depending on the path.
