Question

solve the equation in the complex number system.
$x^4 = 81$
the solution set is {}.
(use a comma to separate answers as needed.)

Explanation:

Step1: Rewrite the equation

Rewrite the equation \(x^{4}=81\) as \(x^{4}-81 = 0\). Notice that this is a difference of squares since \(x^{4}=(x^{2})^{2}\) and \(81 = 9^{2}\). So we can factor it using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\):
\((x^{2}+9)(x^{2}-9)=0\)

Step2: Factor further

The factor \(x^{2}-9\) is also a difference of squares (\(x^{2}-3^{2}\)), so we can factor it again:
\((x^{2}+9)(x + 3)(x - 3)=0\)

Step3: Solve for \(x\) from real factors

First, solve \(x - 3=0\), we get \(x = 3\). Then solve \(x + 3=0\), we get \(x=- 3\).

Step4: Solve for \(x\) from the complex factor

Now solve \(x^{2}+9 = 0\), which can be rewritten as \(x^{2}=-9\). In the complex number system, \(-9 = 9i^{2}\) (since \(i^{2}=-1\)), so \(x^{2}=9i^{2}\). Taking the square root of both sides, we have \(x=\pm3i\).

Answer:

\(3, - 3, 3i, - 3i\)