We need all complex numbers of the form

First condition: $|z-1|\le 1$ means $|(a-1)+ib|\le 1$ so $(a-1)^2+b^2\le 1.$

Since $a,b$ are integers, we find all integer solutions of $(a-1)^2+b^2\le 1.$

Possible values:

• $(a-1)^2=0 \Rightarrow a=1$, then $b^2\le 1 \Rightarrow b=0,\pm 1$
• $(a-1)^2=1 \Rightarrow a=0$ or $2$, then $b^2\le 0 \Rightarrow b=0$

Hence possible lattice points are: $z\in \{1,\;1+i,\;1-i,\;0,\;2\}.$

Second condition: $|z-5|\le |z-5i|.$ Put $z=a+ib$. Then $|z-5|^2=(a-5)^2+b^2,$ $|z-5i|^2=a^2+(b-5)^2.$

So the inequality becomes $(a-5)^2+b^2\le a^2+(b-5)^2.$

Expanding: $a^2-10a+25+b^2\le a^2+b^2-10b+25$ $-10a\le -10b$ $a\ge b.$

Now test the candidate points from step 2:

• $z=1$ gives $(a,b)=(1,0)$, and $1\ge 0$ ✓
• $z=1+i$ gives $(1,1)$, and $1\ge 1$ ✓
• $z=1-i$ gives $(1,-1)$, and $1\ge -1$ ✓
• $z=0$ gives $(0,0)$, and $0\ge 0$ ✓
• $z=2$ gives $(2,0)$, and $2\ge 0$ ✓

So all five points satisfy both conditions.

Now compute the sum of the squares of their moduli.

Recall $|a+ib|^2=a^2+b^2.$

Therefore:

• For $z=1$: $|z|^2=1$
• For $z=1+i$: $|z|^2=1^2+1^2=2$
• For $z=1-i$: $|z|^2=1^2+(-1)^2=2$
• For $z=0$: $|z|^2=0$
• For $z=2$: $|z|^2=4$

Sum: $1+2+2+0+4=9.$

Hence the required integer is $\boxed{9}.$