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

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

4 i \cdot(-7+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 $4i$ and $(-7+i)$ . $\newline$ To multiply two complex numbers, we use the distributive property (also known as the FOIL method for binomials), which states that for any complex numbers $a+bi$ and $c+di$ , $(a+bi)(c+di) = ac + adi + bci + bdi^2$ . $\newline$ Let's apply this to our numbers: $(4i)(-7+i) = (4i)(-7) + (4i)(i)$ .
Apply distributive property: Calculate the individual products. $\newline$ $(4i)(-7) = -28i$ (since $4i \times -7$ gives $-28i$ ). $\newline$ $(4i)(i) = 4i^2$ (since $4i \times i$ gives $4i^2$ ).
Calculate individual products: Remember that $i^2 = -1$ . Substitute $i^2$ with $-1$ in the expression $4i^2$ to get $4(-1)$ , which equals $-4$ .
Substitute $i^2$ with $-1$ : Combine the results from Step $2$ and Step $3$ . $\newline$ We have $-28i$ from the first product and $-4$ from the second product. $\newline$ So, the sum is $-28i + (-4)$ .
Combine results from Step $2$ and Step $3$ : Write the result in the standard form of a complex number $a+bi$ . The real part $a$ is $-4$ , and the imaginary part $b$ is $-28$ . Therefore, the product of the complex numbers $4i$ and $(-7+i)$ is $-4 - 28i$ .

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