Write (-1+2i)^(3) in simplest a+bi form. Answer:

Calculate Square: To solve (-1+2i)^(3), we will first calculate the square of (-1+2i) and then multiply the result by (-1+2i) again. (-1+2i)^(2) = (-1+2i) * (-1+2i) = 1 - 4i + 4i^2. Since i^2 = -1, we replace i^2 with -1. 1 - 4i - 4 = -3 - 4i.Multiply by (-1+2i): Now we have the square of (-1+2i), which is -3 - 4i. We will multiply this result by (-1+2i) to find the cube. (-3 - 4i) * (-1 + 2i) = 3 + 4i - 6i - 8i^2. Again, we replace i^2 with -1. 3 + 4i - 6i + 8 = 11 - 2i.

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Write
(-1+2i)^(3) in simplest
a+bi form.
Answer:

Write $(-1+2 i)^{3}$ in simplest $a+b i$ form. $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

Full solution

Q. Write

(-1+2 i)^{3}

in simplest

a+b i

form.

\newline

Answer:

Calculate Square: To solve $(-1+2i)^{3}$ , we will first calculate the square of $(-1+2i)$ and then multiply the result by $(-1+2i)$ again. $(-1+2i)^{2} = (-1+2i) \times (-1+2i) = 1 - 4i + 4i^{2}$ . Since $i^{2} = -1$ , we replace $i^{2}$ with $-1$ . $1 - 4i - 4 = -3 - 4i$ .
Multiply by $(-1+2i)$ : Now we have the square of $(-1+2i)$ , which is $-3 - 4i$ . We will multiply this result by $(-1+2i)$ to find the cube. $\newline$ $(-3 - 4i) \times (-1 + 2i) = 3 + 4i - 6i - 8i^2.$ $\newline$ Again, we replace $i^2$ with $-1$ . $\newline$ $3 + 4i - 6i + 8 = 11 - 2i.$