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9i*(-4-7i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$9 i \cdot(-4-7 i)=$ $\newline$ 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.

9 i \cdot(-4-7 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Step $1$ : Distribute the multiplication: Multiply the complex numbers $9i$ and $(-4 - 7i)$ . $\newline$ To multiply two complex numbers, we distribute the multiplication over addition, just like we would with binomials. $\newline$ $(9i) * (-4 - 7i) = (9i * -4) + (9i * -7i)$
Step $2$ : Calculate the products: Calculate the products from Step $1$ . $\newline$ $9i \times -4 = -36i$ (since $i$ is the imaginary unit and $-4$ is a real number, their product is an imaginary number). $\newline$ $9i \times -7i = -63i^2$ (since $i^2 = -1$ , we will use this to simplify the expression).
Step $3$ : Simplify using $i^2 = -1$ : Simplify the expression from Step $2$ using the fact that $i^2 = -1$ . $\newline$ $-36i + (-63 \times -1) = -36i + 63$
Step $4$ : Write the final answer: Write the final answer in the form $a + bi$ . $\newline$ The real part $a$ is $63$ , and the imaginary part $b$ is $-36$ . $\newline$ So, the product of the complex numbers $9i$ and $(-4 - 7i)$ is $63 - 36i$ .

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