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

Step-by-step explanation:

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

Then 1 can be written as ...

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

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}

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w

apex answer is: true.