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11*(6i-9)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$11 \cdot(6 i-9)=$ $\newline$ 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.

11 \cdot(6 i-9)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Multiply real part by $11$ : Multiply the real part of the complex number by $11$ . $\newline$ We have the complex number $(6i - 9)$ and we want to multiply it by $11$ . We start by multiplying the real part of the complex number, which is $-9$ , by $11$ . $\newline$ Calculation: $11 \times (-9) = -99$
Multiply imaginary part by $11$ : Multiply the imaginary part of the complex number by $11$ . $\newline$ Next, we multiply the imaginary part of the complex number, which is $6i$ , by $11$ . $\newline$ Calculation: $11 \times (6i) = 66i$
Combine real and imaginary parts: Combine the results from Step $1$ and Step $2$ . $\newline$ We combine the real part and the imaginary part to get the final complex number. $\newline$ Calculation: $(-99) + (66i) = -99 + 66i$

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