Question

# To calculate: i^{1000} + i^{1002}

Factors and multiples
To calculate: The simplified expression for the expression $$\displaystyle{i}^{{{1000}}}+{i}^{{{1002}}}$$

2021-08-11
Formula used:
By the definition of $$\displaystyle{i}=\sqrt{{-{1}}}$$, it follows that:
$$\displaystyle{i}^{{{2}}}=-{1}$$
$$\displaystyle{i}^{{{3}}}=-{i}$$
$$\displaystyle{i}^{{{4}}}={1}$$
To simplify higher powers of i, decompose the expression into multiples of $$\displaystyle{i}^{{{4}}}$$ and write the remaining factors as $$\displaystyle{i},{i}^{{{2}}}$$ or $$\displaystyle{i}^{{{3}}}$$.
Calculation:
Consider the provided expression:
$$\displaystyle{i}^{{{1000}}}+{i}^{{{1002}}}$$
Now, to simplify higher powers of i, decompose the expression into multiples of $$\displaystyle{i}^{{{4}}}{\left({i}^{{{4}}}={1}\right)}$$ and write the remaining factors as $$\displaystyle{i},{i}^{{{2}}}$$ or $$\displaystyle{i}^{{{3}}}$$.
Thus, for the provoded expression:
$$\displaystyle{i}^{{{1000}}}+{i}^{{{1002}}}={i}^{{{4}{\left({250}\right)}}}+{i}^{{{4}{\left({250}\right)}}}\cdot{i}^{{{2}}}$$
$$\displaystyle={1}^{{{250}}}+{1}^{{{250}}}\cdot{i}^{{{2}}}$$
Now, as $$\displaystyle{i}^{{{2}}}=-{1}$$, thus:
$$\displaystyle{i}^{{{1000}}}+{i}^{{{1002}}}={1}^{{{250}}}+{1}^{{{250}}}\cdot{i}^{{{2}}}$$
$$\displaystyle={1}+{\left(-{1}\right)}$$
$$\displaystyle={0}$$
Thus, the simplified expression for the expression $$\displaystyle{i}^{{{1000}}}+{i}^{{{1002}}}$$ is 0.