Let ω be a complex number such that 2ω + 1 = z, where z = √3i. If |||1 1 1 1 - ω² ω² 1 ω² ω 7||| = 3k, then k is equal to.

2ω + 1 = z
2ω + 1 = √3i
ω = (√3i - 1) / 2 ⇒ ω² = ((√3i - 1) / 2)² = (3i² - 2√3i + 1) / 4 = (-3 - 2√3i + 1) / 4 = (-2 - 2√3i) / 4 = (-1 - √3i) / 2 = -ω
Now, |||1 1 1 1 - ω² ω² 1 ω² ω 7||| = 3k
Given ⇒ |||1 1 1 1 ω² ω² 1 ω² ω 7||| = 3k
1(ω² - ω⁴) - (ω - ω²) + 1(ω² - ω) = 3k
3ω² - 3ω = 3k
k = ω² - ω
k = -ω - ω = -2ω
k = -2((√3i - 1) / 2) = 1 - √3i = -z