Question

2i^20 + i^42 i^101 - i^2 Show Work

Answer 1

I assume you're talking about a fraction,

(2i ²⁰ + i ⁴²) / (i ¹⁰¹ - i ²)

Remember that i ² = -1; from this it follows that

i ³ = i ² × i = -i

i ⁴ = i ³ × i = (-i ) × i = -i ² = -(-1) = 1

i ⁵ = i ⁴ × i = i

i ⁶ = i ⁵ × i = i ² = -1

and so on. In particular, raising i to some integer multiple of 4 reduces to 1.

Using the above pattern, we find

i ²⁰ = 1

i ⁴² = i ⁴⁰ × i ² = 1 × (-1) = -1

i ¹⁰¹ = i ¹⁰⁰ × i = 1 × i = i

So, we have

(2i ²⁰ + i ⁴²) / (i ¹⁰¹ - i ²) = (2 - 1) / (i - (-1)) = 1 / (i + 1)

We can stop here, but usually it's more useful to have complex numbers written in standard a + bi form. To get this result, multiply the numerator and denominator by the complex conjugate of i + 1:

(2i ²⁰ + i ⁴²) / (i ¹⁰¹ - i ²) = 1 / (i + 1) × (-i + 1) / (-i + 1)

… = (-i + 1) / (-i ² + 1²)

… = (-i + 1) / (-(-1) + 1)

… = (1 - i ) / 2

or equivalently,

… = 1/2 - 1/2 i