Let $$A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}$$ where $$i = \sqrt{-1}$$. Then, the number of elements in the set $$\{n \in \{1, 2, \ldots, 100\} : A^n = A\}$$ is

We have $$A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}$$ and need to find the number of $$n \in \{1, 2, \ldots, 100\}$$ such that $$A^n = A$$.

To find the eigenvalues of $$A$$, we write the characteristic equation:

$$\det(A - \lambda I) = (1+i-\lambda)(0-\lambda) - (1)(-i) = 0$$

$$-\lambda(1+i-\lambda) + i = 0$$

$$\lambda^2 - (1+i)\lambda + i = 0$$

Using the quadratic formula:

$$\lambda = \frac{(1+i) \pm \sqrt{(1+i)^2 - 4i}}{2} = \frac{(1+i) \pm \sqrt{2i - 4i}}{2} = \frac{(1+i) \pm \sqrt{-2i}}{2}$$

Now $$-2i = 2e^{-i\pi/2}$$, so $$\sqrt{-2i} = \sqrt{2}\,e^{-i\pi/4} = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$$.

$$\lambda_1 = \frac{(1+i) + (1-i)}{2} = 1, \qquad \lambda_2 = \frac{(1+i) - (1-i)}{2} = i$$

Next, since the eigenvalues $$\lambda_1 = 1$$ and $$\lambda_2 = i$$ are distinct, $$A$$ is diagonalizable:

$$A = PDP^{-1}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$$

Therefore, $$A^n = PD^nP^{-1}$$, where $$D^n = \begin{pmatrix} 1 & 0 \\ 0 & i^n \end{pmatrix}$$.

In order for $$A^n = A$$, we require $$D^n = D$$, which means:

$$1^n = 1 \quad \text{(always true)}, \quad \text{and} \quad i^n = i$$

$$i^n = i$$ holds when $$n \equiv 1 \pmod{4}$$.

Finally, we count the valid values of $$n$$.

We need $$n \in \{1, 2, \ldots, 100\}$$ with $$n \equiv 1 \pmod{4}$$.

These are: $$n = 1, 5, 9, 13, \ldots, 97$$.

This is an arithmetic sequence with first term 1, common difference 4, and last term 97.

Number of terms: $$\frac{97 - 1}{4} + 1 = 25$$.

The answer is $$\boxed{25}$$.