Consider the equation: z^(4)=16 One solution is z=2, another is -2 , and there are 2 more distinct complex solutions. What are those solutions? Choose 2 answers: A z=4i B z=2i c. z=-4i D z=-2i

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Consider the equation:  z^(4)=16 One solution is  z=2, another is -2 , and there are 2 more distinct complex solutions. What are those solutions? Choose 2 answers: A  z=4i B  z=2i c.  z=-4i D  z=-2i

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Identify type of solutions: Write down the given equation and identify the type of solutions we are looking for. The given equation is z^(4) = 16. We know that z = 2 and z = -2 are real solutions to this equation. Since the equation is a polynomial of degree 4, it must have 4 solutions in total (counting multiplicity) according to the Fundamental Theorem of Algebra. We are looking for the two complex solutions that are not real.Express equation in complex form: Express the equation in its complex form. The equation z^(4) = 16 can be rewritten as z^(4) - 16 = 0. This is a difference of squares and can be factored as (z^2 - 4)(z^2 + 4) = 0.Solve for z when z^2 - 4 = 0: Solve for z when z^2 - 4 = 0. We already know the solutions for z^2 - 4 = 0 are z = 2 and z = -2. These are the real solutions.Solve for z when z^2 + 4 = 0: Solve for z when z^2 + 4 = 0. To find the complex solutions, we solve the equation z^2 + 4 = 0. This gives us z^2 = -4. Taking the square root of both sides, we get z = ±√(-4).Calculate square root of -4: Calculate the square root of -4. The square root of -4 is 2i, where i is the imaginary unit. Therefore, the two complex solutions are z = 2i and z = -2i.Match solutions with options: Match the solutions with the given options. The solutions we found are z = 2i and z = -2i. These correspond to options B and D.

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