4. Given four non-zero complex numbers $a, b, c, d$ satisfying $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$. Then the number of different values that $\frac{a-b+c+d}{a+b-c+d}$ can take is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
4. (D).

Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then $a=b k=c k^{2}=d k^{3}=a k^{4}$.
That is $\square$
$$
\begin{array}{l}
b=a k^{3}, c=a k^{2}, d=a k, a=u k^{4} . \\
\therefore k^{4}=1 . \text { Hence } k= \pm 1, \pm \mathrm{i} . \\
\therefore \frac{a}{a} \mp \frac{b+c+d}{b-c+d}=\frac{a+(-b)+c+d}{b+(-c)+d+a} \\
\quad=k= \pm 1 \text { or } \pm \mathrm{i} .
\end{array}
$$