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vladimir2022 [97]
3 years ago
9

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

Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

I assume you're talking about a fraction,

(2<em>i</em> ²⁰ + <em>i</em> ⁴²) / (<em>i</em> ¹⁰¹ - <em>i</em> ²)

Remember that <em>i</em> ² = -1; from this it follows that

<em>i</em> ³ = <em>i</em> ² × <em>i</em> = -<em>i</em>

<em>i</em> ⁴ = <em>i</em> ³ × <em>i</em> = (-<em>i </em>) × <em>i</em> = -<em>i </em>² = -(-1) = 1

<em>i</em> ⁵ = <em>i</em> ⁴ × <em>i</em> = <em>i</em>

<em>i</em> ⁶ = <em>i</em> ⁵ × <em>i</em> = <em>i</em> ² = -1

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

Using the above pattern, we find

<em>i</em> ²⁰ = 1

<em>i</em> ⁴² = <em>i</em> ⁴⁰ × <em>i</em> ² = 1 × (-1) = -1

<em>i</em> ¹⁰¹ = <em>i</em> ¹⁰⁰ × <em>i</em> = 1 × <em>i</em> = <em>i</em>

So, we have

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

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

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

… = (-<em>i</em> + 1) / (-<em>i</em> ² + 1²)

… = (-<em>i</em> + 1) / (-(-1) + 1)

… = (1 - <em>i</em> ) / 2

or equivalently,

… = 1/2 - 1/2 <em>i</em>

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