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3i*(3+2i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$3 i \cdot(3+2 i)=$ $\newline$ 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.

3 i \cdot(3+2 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Multiply complex numbers: Multiply the complex numbers $3i$ and $(3+2i)$ . $\newline$ To multiply two complex numbers, we distribute the multiplication over addition, just like we would with binomials. $\newline$ $(3i)\cdot(3+2i) = 3i\cdot 3 + 3i\cdot 2i$
Calculate the products: Calculate the products. $\newline$ $3i \times 3 = 9i$ (since $i$ is the imaginary unit, it stays with the real number $3$ ) $\newline$ $3i \times 2i = 6i^2$ (since $i \times i = i^2$ , and $i^2 = -1$ )
Substitute and simplify: Substitute $i^2$ with $-1$ and simplify. $\newline$ $6i^2 = 6*(-1) = -6$ (since $i^2 = -1$ ) $\newline$ Now we combine this with the other product: $\newline$ $9i + (-6)$
Combine the products: Write the final answer in the form $a+bi$ . $\newline$ The real part $a$ is $-6$ , and the imaginary part $b$ is $9$ . $\newline$ So the final answer is $-6 + 9i$ .

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