i) $(r-1)'s$ complement of $N =r^n- r^{-m}-N$
$n$ is no of digits in integer part and $m$ is no of digits in fractional part.
so, for $m=0$
$(r-1)'s$ complement of $N =r^n- 1-N$
$r^n- 1 =$ biggest $n$ digit number in base $r$
for example, n=4
for $r= 10$, $10^4- 1=(9999)_{10}$
for $r=2$, $2^4- 1=(1111)_2$
for $r=8$, $8^4- 1=(7777)_8$
for $r=16$, $16^4- 1=(FFFF)_{16}$
ii) $(r-1)'s$ complement of $N = r^n-1-N$
$r's$ complement of $N = r^n-N$
that is, also we can say $r's$ complement of $N = (r-1)'s$ complement $+1$.
in both cases, result will same, it's all your choice.
if we use $r^n-N$, we need to subtract N from smallest $n+1$ digit number ($r^n$) in base r.
if we use $(r-1)'s$ complement $+1$, then for $r^n-1-N$, we need to subtract N from largest $n$ digit number ($r^n-1$) in base r.
Second way is fast in terms of calculations as I feel.