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11 i*(-8+10 i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$11 i \cdot(-8+10 i)=$ $\newline$ 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.

11 i \cdot(-8+10 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Write down complex numbers: Write down the complex numbers to be multiplied. $\newline$ We have two complex numbers: $11i$ and $(-8+10i)$ . We need to multiply these two complex numbers together.
Use distributive property: Use the distributive property to multiply the complex numbers. $\newline$ Multiplying $11i$ by each term in the complex number $(-8+10i)$ gives us: $\newline$ $11i \cdot (-8) + 11i \cdot (10i)$
Calculate the products: Calculate the products. $\newline$ $11i \times (-8) = -88i$ (since $i$ is the imaginary unit and $-8$ is a real number) $\newline$ $11i \times (10i) = 110i^2$ (since $i \times i = i^2$ )
Simplify the expression: Simplify the expression. $\newline$ We know that $i^2 = -1$ , so we can replace $i^2$ with $-1$ in the expression: $\newline$ $-88i + 110(-1)$ $\newline$ This simplifies to: $\newline$ $-88i - 110$
Write final answer: Write the final answer in the form $a+bi$ . The real part $a$ is $-110$ , and the imaginary part $b$ is $-88$ . So the product of $11i$ and $(-8+10i)$ is: $-110 - 88i$

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