Write (-1+4i)^(3) in simplest a+bi form. Answer:

Recognize Complex Number: To solve (-1+4i)^(3), we first need to recognize that we are raising a complex number to the third power. We can do this by multiplying the complex number by itself three times.Square Intermediate Result: First, let's square (-1+4i) to get an intermediate result. We use the formula (a+bi)(a+bi) = a^2 + 2abi - b^2 (since i^2 = -1). So, (-1+4i)^2 = (-1)^2 + 2*(-1)*4i + (4i)^2 = 1 - 8i - 16 (since i^2 = -1) = -15 - 8i.Multiply to Find Result: Now, we need to multiply our result by (-1+4i) again to find (-1+4i)^3. So, (-15 - 8i)(-1 + 4i) = (-15)*(-1) + (-15)*4i + (-8i)*(-1) + (-8i)*(4i) = 15 - 60i + 8i - 32 (since i^2 = -1) = 15 - 52i - 32 = -17 - 52i.

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Write
(-1+4i)^(3) in simplest
a+bi form.
Answer:

Write $(-1+4 i)^{3}$ in simplest $a+b i$ form. $\newline$ Answer:

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Precalculus

Find the roots of factored polynomials

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Q. Write

(-1+4 i)^{3}

in simplest

a+b i

form.

\newline

Answer:

Recognize Complex Number: To solve $(-1+4i)^{3}$ , we first need to recognize that we are raising a complex number to the third power. We can do this by multiplying the complex number by itself three times.
Square Intermediate Result: First, let's square $(-1+4i)$ to get an intermediate result. We use the formula $(a+bi)(a+bi) = a^2 + 2abi - b^2$ (since $i^2 = -1$ ). $\newline$ So, (-1+4i)^2 = (-1)^2 + 2*(-1)*4i + (4i)^2$\newline= 1 - 8i - 16$ (since $i^2 = -1$ ) $\newline$ = $-15 - 8i$ .
Multiply to Find Result: Now, we need to multiply our result by $(-1+4i)$ again to find $(-1+4i)^3$ . So, $(-15 - 8i)(-1 + 4i) = (-15)\cdot(-1) + (-15)\cdot4i + (-8i)\cdot(-1) + (-8i)\cdot(4i) = 15 - 60i + 8i - 32$ (since $i^2 = -1$ ) $= 15 - 52i - 32 = -17 - 52i$ .