Answer:
a. We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution.
b. (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
Step-by-step explanation:
a. Why do we need imaginary numbers?
We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution. For example, the equation of the form x² + 2x + 1 = 0 has the solution (x - 1)(x + 1) = 0 , x = 1 twice. The equation x² + 1 = 0 has the solution x² = -1 ⇒ x = √-1. Since we cannot find the square-root of a negative number, the identity i = √-1 was developed to be the solution to the problem of solving quadratic equations which have the square-root of a negative number.
b. Expand (a + ib)⁵
(a + ib)⁵ = (a + ib)(a + ib)⁴ = (a + ib)(a + ib)²(a + ib)²
(a + ib)² = (a + ib)(a + ib) = a² + 2iab + (ib)² = a² + 2iab - b²
(a + ib)²(a + ib)² = (a² + 2iab - b²)(a² + 2iab - b²)
= a⁴ + 2ia³b - a²b² + 2ia³b + (2iab)² - 2iab³ - a²b² - 2iab³ + b⁴
= a⁴ + 2ia³b - a²b² + 2ia³b - 4a²b² - 2iab³ - a²b² - 2iab³ + b⁴
collecting like terms, we have
= a⁴ + 2ia³b + 2ia³b - a²b² - 4a²b² - a²b² - 2iab³ - 2iab³ + b⁴
= a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴
(a + ib)(a + ib)⁴ = (a + ib)(a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴)
= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b + 4i²a³b² - 6ia²b³ - 4i²ab⁴ + ib⁵
= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b - 4a³b² - 6ia²b³ + 4ab⁴ + ib⁵
collecting like terms, we have
= a⁵ + 4ia⁴b + ia⁴b - 6a³b² - 4a³b² - 4ia²b³ - 6ia²b³ + ab⁴ + 4ab⁴ + ib⁵
= a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
So, (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵
, where B is the base and h is the height. We can rearrange this formula to solve for B.
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