1. **State the problem:** Simplify and evaluate the expression $$\left( \frac{1}{1+i} - \frac{1}{1-i} \right)^{100} = \left( \frac{2+3w}{2w^{2}+3} + \frac{4w^{2}+1}{4+w} \right)^{200}$$ for complex numbers $i$ and $w$. 2. **Simplify the left side:** Calculate $$\frac{1}{1+i} - \frac{1}{1-i}$$. Use the conjugate to rationalize denominators: $$\frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1 - i}{1 - i^{2}} = \frac{1 - i}{1 - (-1)} = \frac{1 - i}{2}$$ Similarly, $$\frac{1}{1 - i} = \frac{1 + i}{(1 - i)(1 + i)} = \frac{1 + i}{1 - (-1)} = \frac{1 + i}{2}$$ So, $$\frac{1}{1+i} - \frac{1}{1-i} = \frac{1 - i}{2} - \frac{1 + i}{2} = \frac{(1 - i) - (1 + i)}{2} = \frac{1 - i - 1 - i}{2} = \frac{-2i}{2} = -i$$ 3. **Raise to the 100th power:** $$(-i)^{100} = \left((-i)^4\right)^{25}$$ Since $$(-i)^4 = ((-1) \cdot i)^4 = (-1)^4 \cdot i^4 = 1 \cdot 1 = 1$$ Therefore, $$(-i)^{100} = 1^{25} = 1$$ 4. **Simplify the right side:** $$\frac{2+3w}{2w^{2}+3} + \frac{4w^{2}+1}{4+w}$$ Without specific value or relation for $w$, we cannot simplify further. The problem likely implies equality holds for some $w$ or is an identity. 5. **Summary:** The left side simplifies to 1, so the equation becomes: $$1 = \left( \frac{2+3w}{2w^{2}+3} + \frac{4w^{2}+1}{4+w} \right)^{200}$$ This implies: $$\frac{2+3w}{2w^{2}+3} + \frac{4w^{2}+1}{4+w} = 1$$ if the right side is real and positive. **Final answer:** $$\boxed{1}$$