Question

Please help me with this math problem. I don't get it and need help to complete it!

For each odd positive integer n, the only real number solution to x^n=1 is x = 1 while for even positive integers n, x=1 and x= − 1 are solutions to x^n=1x n = 1. In this problem we look for all complex number solutions to LaTeX: x^n=1 for some small values of n.


1. Find all complex numbers a + bi whose cube is 1.

2.Find all complex numbers a + bi whose fourth power is 1.

Answer 1

Answer:

1. 1, cos(2π/3)+i·sin(2π/3), cos(4π/3) +i·sin(4π/3)
2. 1, i, -1, -i

Step-by-step explanation:

Euler's formula comes into play here. That tells you ...

  $e^{ix}=\cos{x}+i\sin{x}$

Then 1 can be written as ...

  $1 = e^{2ki\pi} \qquad\text{for any integer k}$

Then the n-th root of 1 will be ...

  $1^{\frac{1}{n}}=e^{2ki\pi /n}=\cos{(2k\pi /n)}+i\sin{(2k\pi /n)}$

Since the trig functions are periodic with period 2π, useful values of k are from 0 to n-1.

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1) The cube roots of 1 are ...

  cos(2πk/3) +i·sin(2πk/3) . . . . for k = 0 to 2

  = {1, cos(2π/3) +i·sin(2π/3), cos(4π/3) +i·sin(4π/3)}

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2) The fourth roots of 1 are ...

  cos(2πk/4) +i·sin(2πk/4) = cos(kπ/2) +i·sin(kπ/2)

  = {1, i, -1, -i}