3. Let $z$ be a complex number, satisfying $\left|z-\frac{1}{z}\right|=1$. Then the sum of the maximum and minimum values of $|z|$ is ( ).
(A) $-2+2 \sqrt{5}$
(B) $\frac{3+\sqrt{5}}{2}$
(C) $\sqrt{5}$
(D) A different value from the above three options.
3.C.

Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, then
$$
\begin{aligned}
1 & =\left|z-\frac{1}{z}\right|=\left|x+y \mathrm{i}-\frac{x-y \mathrm{i}}{x^{2}+y^{2}}\right| \\
& =\sqrt{\left(x-\frac{x}{x^{2}+y^{2}}\right)^{2}+\left(y+\frac{y}{x^{2}+y^{2}}\right)^{2}} \\
& =\frac{\sqrt{x^{2}\left(x^{2}+y^{2}-1\right)^{2}+y^{2}\left(x^{2}+y^{2}+1\right)^{2}}}{x^{2}+y^{2}} .
\end{aligned}
$$

Thus, $x^{2}\left(x^{2}+y^{2}-1\right)^{2}+y^{2}\left(x^{2}+y^{2}+1\right)^{2}=\left(x^{2}+y^{2}\right)^{2}$,
which simplifies to $\left(x^{2}+y^{2}\right)^{2}-3 x^{2}+y^{2}+1=0$.
Let $x^{2}+y^{2}=t^{2}(t \geqslant 0)$, then $y^{2}=t^{2}-x^{2}$, and $x^{2} \leqslant t^{2}$.
Substituting into the equation (1) gives $t^{4}+t^{2}+1=4 x^{2} \leqslant 4 t^{2}$, which simplifies to
$$
t^{4}-3 t^{2}+1 \leqslant 0 \text {. }
$$

Solving this, we get $\frac{3-\sqrt{5}}{2} \leqslant t^{2} \leqslant \frac{3+\sqrt{5}}{2}$, which means $\frac{-1+\sqrt{5}}{2} \leqslant t \leqslant \frac{1+\sqrt{5}}{2}$. Therefore, $\frac{-1+\sqrt{5}}{2} \leqslant|z|=t \leqslant \frac{1+\sqrt{5}}{2}$, with equality holding if and only if $x^{2}=t^{2}$, i.e., $y=0$ and $|x|=t=\frac{-1+\sqrt{5}}{2}$ or $\frac{1+\sqrt{5}}{2}$.
Thus, $|z|_{\max }+|z|_{\min }=\frac{-1+\sqrt{5}}{2}+\frac{1+\sqrt{5}}{2}=\sqrt{5}$.