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

Expand Expression: Expand (1-3i)^(3) using the binomial theorem or by multiplying (1-3i) by itself three times. (1-3i) * (1-3i) * (1-3i) = (1-3i) * ((1-3i) * (1-3i)) First, we'll multiply the first two factors: (1-3i) * (1-3i) = 1^2 - 2*1*3i + (3i)^2 = 1 - 6i - 9i^2 Since i^2 = -1, we replace i^2 with -1: = 1 - 6i - 9(-1) = 1 - 6i + 9 = 10 - 6iFirst Multiplication: Now, multiply the result from Step 1 by (1-3i): (10 - 6i) * (1-3i) = 10(1) - 10(3i) - 6i(1) + 6i(3i) = 10 - 30i - 6i + 18i^2 Again, replace i^2 with -1: = 10 - 30i - 6i + 18(-1) = 10 - 30i - 6i - 18 Combine like terms: = (10 - 18) - (30i + 6i) = -8 - 36i

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

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Math Problems

Algebra 2

Csc, sec, and cot of special angles

Full solution

Q. Write

(1-3 i)^{3}

in simplest

a+b i

form.

\newline

Answer:

Expand Expression: Expand $(1-3i)^{3}$ using the binomial theorem or by multiplying $(1-3i)$ by itself three times. $\newline$ (1-3i) \times (1-3i) \times (1-3i) = (1-3i) \times ((1-3i) \times (1-3i))$\newlineFirst, we'll multiply the first two factors: $\newline$ \$(1-3i) \times (1-3i) = 1^{2} - 2\times1\times3i + (3i)^{2}\(\newline$= 1 - 6i - 9i^{2}$\newline$Since \$i^{2} = -1\), we replace $i^{2}$ with $-1$:$\newline$= $1$ - $6$i - $9$($-1$)$\newline$= $1$ - $6$i + $9$$\newline$= $10$ - $6$i
First Multiplication: Now, multiply the result from Step $1$ by $(1-3i)$:$\newline$$(10 - 6i) \times (1-3i) = 10(1) - 10(3i) - 6i(1) + 6i(3i)$$\newline$$= 10 - 30i - 6i + 18i^2$$\newline$Again, replace $i^2$ with $-1$:$\newline$$= 10 - 30i - 6i + 18(-1)$$\newline$$= 10 - 30i - 6i - 18$$\newline$Combine like terms:$\newline$$= (10 - 18) - (30i + 6i)$$\newline$$= -8 - 36i$