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

Calculate (-1+3i)^(2): To find (-1+3i)^(3), we first need to calculate (-1+3i)^(2). This is done by multiplying the complex number by itself. (-1+3i) * (-1+3i) = 1 - 3i - 3i + 9i^2 Since i^2 = -1, we can simplify this to: 1 - 6i - 9(-1) = 1 - 6i + 9 Combining like terms, we get: 10 - 6iMultiply by (-1+3i): Now we need to multiply the result from the first step by (-1+3i) to get (-1+3i)^(3). (10 - 6i) * (-1 + 3i) = -10 + 30i + 6i - 18i^2 Again, using i^2 = -1, we can simplify this to: -10 + 30i + 6i + 18 Combining like terms, we get: 8 + 36i

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

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

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Precalculus

Find the roots of factored polynomials

Full solution

Q. Write

(-1+3 i)^{3}

in simplest

a+b i

form.

\newline

Answer:

Calculate $(-1+3i)^{2}$ : To find $(-1+3i)^{3}$ , we first need to calculate $(-1+3i)^{2}$ . This is done by multiplying the complex number by itself. $\newline$ $(-1+3i) \times (-1+3i) = 1 - 3i - 3i + 9i^2$ $\newline$ Since $i^2 = -1$ , we can simplify this to: $\newline$ $1 - 6i - 9(-1) = 1 - 6i + 9$ $\newline$ Combining like terms, we get: $\newline$ $10 - 6i$
Multiply by $(-1+3i)$ : Now we need to multiply the result from the first step by $(-1+3i)$ to get $(-1+3i)^{3}$ . $(10 - 6i) * (-1 + 3i) = -10 + 30i + 6i - 18i^2$ Again, using $i^2 = -1$ , we can simplify this to: $-10 + 30i + 6i + 18$ Combining like terms, we get: $8 + 36i$