5. Let $n \in \mathbf{N}, r>0$. Then the number of complex roots of the equation
$$
x^{n+1}+r x^{n}-r^{n+1}=0
$$
with modulus $r$ is ( ).
(A) 0
(B) 1
(C) 2
(D) greater than 2
5.A.

The equation is transformed into
$$
x^{n}(x+r)=r^{n+1} \text {. }
$$

Since $|x|=r$, taking the modulus of both sides of equation (1) yields $|x+r|=r$. The complex roots of the original equation with modulus $r$ are the complex numbers corresponding to the intersection points of the two circles $|x|=r$ and $|x+r|=r$:
$$
\begin{array}{l}
x_{1}=r\left(\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}\right), \\
x_{2}=r\left(\cos \frac{4 \pi}{3}+\mathrm{i} \sin \frac{4 \pi}{3}\right) .
\end{array}
$$

Substituting (2) into the original equation and simplifying, we get
$$
\begin{array}{l}
\cos \frac{(2 n+1) \pi}{3}+\mathrm{i} \sin \frac{(2 n+1) \pi}{3}=1 \\
\Rightarrow \cos \frac{(2 n+1) \pi}{3}=1, \sin \frac{(2 n+1) \pi}{3}=0 \\
\Rightarrow \frac{(2 n+1) \pi}{3}=2 k \pi(k \in Z), \text { i.e., } 2 n+1=6 k, \text { a }
\end{array}
$$

contradiction.
Similarly, $x_{2}$ does not satisfy the original equation.
Therefore, the original equation has no complex roots with modulus $r$.