1. Analyze Condition (I)
The first condition is given by:
$$1 + \omega^n + \omega^{2n} = 0$$
Here, $$\omega$$ is a primitive cube root of unity, which means $$\omega^3 = 1$$ and $$1 + \omega + \omega^2 = 0$$.
If $$n$$ is a multiple of $$3$$ ($$n = 3k$$), then $$\omega^n = 1$$ and $$\omega^{2n} = 1$$. The sum becomes $$1 + 1 + 1 = 3 \neq 0$$.
If $$n$$ is not a multiple of $$3$$ ($$n \neq 3k$$), then $$\omega^n$$ and $$\omega^{2n}$$ are equal to $$\omega$$ and $$\omega^2$$ in some order. The sum becomes $$1 + \omega + \omega^2 = 0$$.
Therefore, Condition (I) is satisfied if and only if $$n$$ is **not a multiple of 3**.
$$n \not\equiv 0 \pmod 3$$
2. Analyze Condition (II)
The second condition is given by:
$$\left(\frac{1+i}{1-i}\right)^n = 1$$
First, simplify the base complex fraction inside the parenthesis by multiplying the numerator and denominator by the conjugate $$(1+i)$$:
$$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2}$$
Since $$i^2 = -1$$, this simplifies to:
$$\frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$$
Now substitute this back into Condition (II):
$$i^n = 1$$
The powers of $$i$$ repeat in a cycle of $$4$$ ($$i^1=i, i^2=-1, i^3=-i, i^4=1$$). Therefore, $$i^n = 1$$ is satisfied if and only if $$n$$ is a **multiple of 4**.
$$n \equiv 0 \pmod 4$$
3. Combine Both Conditions
We need to find the number of integers in the set $$\{1, 2, 3, \ldots, 300\}$$ that are multiples of $$4$$ but **not** multiples of $$3$$.
Find the total number of multiples of $$4$$ up to $$300$$:
$$\text{Total multiples of 4} = \left\lfloor \frac{300}{4} \right\rfloor = 75$$
Find the number of multiples of $$4$$ that are also multiples of $$3$$ (which means they are multiples of $$\text{LCM}(3,4) = 12$$):
$$\text{Multiples of 12} = \left\lfloor \frac{300}{12} \right\rfloor = 25$$
Subtract the overlapping multiples of $$12$$ from the total multiples of $$4$$:
$$\text{Favorable integers} = 75 - 25 = 50$$
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