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-6i*(6-2i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$-6 i \cdot(6-2 i)=$ $\newline$ Your answer should be a complex number in the form $a+b i$ where $a$ and $b$ are real numbers.

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Write equations of cosine functions using properties

Full solution

Q.

-6 i \cdot(6-2 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Write Problem: Write down the multiplication problem. $\newline$ We need to multiply the complex numbers $-6i$ and $(6-2i)$ .
Distribute $-6i$ : Distribute $-6i$ across the terms in the parentheses. $\newline$ $-6i \times 6 + (-6i) \times (-2i)$
Multiply Real and Imaginary: Multiply the real part by the imaginary part. $\newline$ $-6i \times 6 = -36i$
Simplify Multiplication: Multiply the imaginary parts together. $\newline$ $(-6i) \times (-2i) = 12i^2$
Remember $i^2$ : Remember that $i^2$ is equal to $-1$ . $12i^2 = 12 \times (-1)$
Combine Parts: Simplify the multiplication. $\newline$ $12 \times (-1) = -12$
Combine Parts: Simplify the multiplication. $\newline$ $12 \times (-1) = -12$ Combine the real and imaginary parts. $\newline$ $-36i + (-12) = -12 - 36i$

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