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13 i*(1-5i)=
Your answer should be a complex number in the form
a+bi where
a and
b are real numbers.

$13 i \cdot(1-5 i)=$ $\newline$ 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.

13 i \cdot(1-5 i)=

\newline

Your answer should be a complex number in the form

a+b i

where

a

and

b

are real numbers.

Write down multiplication: Write down the multiplication of the two complex numbers. $\newline$ We are given the complex numbers $13i$ and $(1-5i)$ , and we need to find their product.
Distribute first complex number: Distribute the first complex number over the second. $\newline$ To multiply these two complex numbers, we use the distributive property $(a+bi)(c+di) = ac + adi + bci + bdi^2$ . $\newline$ So, $13i*(1-5i) = 13i\cdot 1 - 13i\cdot 5i$ .
Perform the multiplication: Perform the multiplication. $\newline$ Now we multiply the terms: $\newline$ $13i \times 1 = 13i$ (since anything times $1$ is itself), $\newline$ and $-13i \times 5i = -65i^2$ (since we multiply the coefficients and the imaginary units).
Remember $i^2 = -1$ : Remember that $i^2 = -1$ . $\newline$ The imaginary unit $i$ has the property that $i^2 = -1$ . We use this to simplify the term $-65i^2$ . $\newline$ $-65i^2 = -65*(-1) = 65$ .
Combine real and imaginary parts: Combine the real and imaginary parts. $\newline$ Now we add the real part from step $4$ and the imaginary part from step $3$ to get the final answer. $\newline$ Real part: $65$ (from $-65i^2$ ), $\newline$ Imaginary part: $13i$ (from $13i \cdot 1$ ). $\newline$ So, the product is $65 + 13i$ .

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\newline

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