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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

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

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Solve rational equations

Full solution

Q. Consider the equation:

\newline

z^{4}=16

\newline

One solution is

z=2

, another is

-2

, and there are

2

more distinct complex solutions.

\newline

What are those solutions?

\newline

Choose

\mathbf{2}

answers:

\newline

A

z=4 i

\newline

B

z=2 i

\newline

c.

z=-4 i

\newline

D

z=-2 i

Identify type of solutions: Write down the given equation and identify the type of solutions we are looking for. $\newline$ 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. $\newline$ 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 $z = \pm\sqrt{-4}$ .
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. $\newline$ The solutions we found are $z = 2i$ and $z = -2i$ . These correspond to options $B$ and $D$ .

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