How to plot root loci in z plane
As @TimWescott points out, it doesn't matter if it's s-plane, z-plane, or any other-plane, the problem is one of finding the roots of a polynomial. In this particular case, however,the problem is easily solved without recourse to classical root locus analysis.

Generally, we can express the case of three repeated roots as solving the characteristic equation:

$$\small (z-\alpha)^3+K=0$$

where \$\small K\ge 0\$ is the loop gain.

This is conveniently achieved by solving for the roots of:

$$\small (z-\alpha)=(-K)^{1/3}=a+jb$$

that leads to the root locus: 
$$\small z=\alpha+(-K)^{1/3}=\alpha +a+jb$$

Thus:

$$\small (a+jb)^3=-K$$
Expanding:
$$\small(a^3-3ab^2)+j(3a^2b-b^3)=-K $$

Equating the imaginary term to zero:
$$\small b(3a^2-b^2)=0$$

which reveals the three solutions:
$$\small b=0; \:\:b=\pm ja\sqrt 3$$
Equating the real term to \$\small -K\$ and setting \$\small b=0\$:
$$\small a=(-K)^{1/3} $$

Thus, \$\small a\$ is negative.

The three loci are, therefore:

(1)  \$\small z=\alpha +a \$, which starts at \$\small z=\alpha\$ and travels to \$\small -\infty\$ along the real axis

(2)  \$\small z=\alpha +a +ja\sqrt{3}\$, which starts at \$\small z=\alpha\$ and travels to \$\small \infty\$ along a line at \$\small -60^o\$ to the real axis

(3)   \$\small z=\alpha +a -ja\sqrt{3}\$, which starts at \$\small z=\alpha\$ and travels to \$\small \infty\$ along a line at \$\small 60^o\$ to the real axis