Question

Which expression represents a 5th root of –i?
cos(7pie/4)+sin(7pie/4)
cos(7pie/6)+sin(7pie/6)
cos(7pie/8)+sin(7pie/8)
cos(7pie/10)+sin(7pie/10)

Answer 1

Use DeMoivre's formula:

(cos(θ) + i sin(θ))ⁿ = cos(n θ) + i sin(n θ)

If a given number is a 5th root of -i, raising it to the 5th power should return -i.

Since -i = cos(-π/2) + i sin(-π/2), you should check any number with an argument θ such that 5θ = -π/2. This happens only for the last choice:

(cos(7π/10) + i sin(7π/10))⁵ = cos(35π/10) + i sin(35π/10)

and 35π/10 is equivalent to -π/2 (modulo 2π).

Answer 2

The 5th root of -i is (d) $\mathbf{(cos(\frac{7\pi}{10}) + i sin(\frac{7\pi}{10}))}$

By DeMoivre's formula, we have:

$\mathbf{(cos(\theta) + i sin(\theta))^n = cos(n \theta) + i sin(n \theta)}$

The 5th root of -i means that: n = 5.

So, we have:

$\mathbf{(cos(\theta) + i sin(\theta))^5 = cos(5 \theta) + i sin(5 \theta)}$

Set $\mathbf{\theta}$ to any expression such that:

$\mathbf{\theta = \frac{7\pi}{10}}$

So, we have:

$\mathbf{(cos(\frac{7\pi}{10}) + i sin(\frac{7\pi}{10}))^5 = cos(5 \times \frac{7\pi}{10}) + i sin(5 \times \frac{7\pi}{10})}$

Expand

$\mathbf{(cos(\frac{7\pi}{10}) + i sin(\frac{7\pi}{10}))^5 = cos(\frac{35\pi}{10}) + i sin(\frac{35\pi}{10})}$

Take 5th roots of both sides

$\mathbf{(cos(\frac{7\pi}{10}) + i sin(\frac{7\pi}{10})) = \sqrt[5]{cos(\frac{35\pi}{10}) + i sin(\frac{35\pi}{10})}}$

This means that:

$\mathbf{(cos(\frac{7\pi}{10}) + i sin(\frac{7\pi}{10}))}$ is the 5th root of -i

Hence, option (d) is correct

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