Consider the equation: z^(4)=81 One solution is z=3, another is -3 , and there are 2 more distinct complex solutions. What are those solutions? Choose 2 answers: A z=3i B z=27 i c. z=-3i D z=-27 i

Question

Consider the equation:  z^(4)=81 One solution is  z=3, another is -3 , and there are 2 more distinct complex solutions. What are those solutions? Choose  2 answers: A  z=3i B  z=27 i c.  z=-3i D  z=-27 i

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Rewrite as Difference of Squares: Recognize that the equation z^(4)=81 can be rewritten as z^(4) - 81 = 0, which is a difference of squares.Factor using Formula: Factor the equation using the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b). In this case, we have z^(4) - 81 = (z^2 + 9)(z^2 - 9).Further Factorization: Notice that z^2 - 9 is also a difference of squares, so we can factor it further to (z + 3)(z - 3). Now we have (z^2 + 9)(z + 3)(z - 3) = 0.Find Solutions for z^2 + 9: We already know that z = 3 and z = -3 are solutions from the factors (z - 3) and (z + 3). Now we need to find the solutions for z^2 + 9 = 0.Solve for z: Solve the equation z^2 + 9 = 0 for z. Subtract 9 from both sides to get z^2 = -9.Complex Solutions: Take the square root of both sides to solve for z. The square root of -9 is 3i and -3i, so z = 3i and z = -3i are the complex solutions.Check Validity: Check the solutions by substituting them back into the original equation z^(4)=81. For z = 3i, (3i)^4 = 81i^4 = 81(-1)^2 = 81, which is true. For z = -3i, (-3i)^4 = 81i^4 = 81(-1)^2 = 81, which is also true.

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