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If $(3-2i)(3-i)(2+i)=a+ib$ , where $a$ and $b$ are real numbers and the imaginary number $i$ is such that $i^2=-1$ , what is the value of $(2a+3b)$ ?

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Describe the graph of a linear equation

Full solution

Q. If

(3-2i)(3-i)(2+i)=a+ib

, where

a

and

b

are real numbers and the imaginary number

i

is such that

i^2=-1

, what is the value of

(2a+3b)

?

Multiply Complex Numbers: Step $1$ : Multiply the first two complex numbers $(3−2i)$ and $(3−i)$ . $\newline$ Calculation: $(3−2i)(3−i) = 3(3) + 3(-i) - 2i(3) - 2i(-i) = 9 - 3i - 6i + 2i^2 = 9 - 9i + 2(-1) = 7 - 9i$ .
Multiply by Third Number: Step $2$ : Multiply the result by the third complex number $(2+i)$ . $\newline$ Calculation: $(7 - 9i)(2 + i) = 7(2) + 7(i) - 9i(2) - 9i(i) = 14 + 7i - 18i - 9i^2 = 14 - 11i - 9(-1) = 23 - 11i$ .
Identify Real and Imaginary Parts: Step $3$ : Identify the real part $a$ and the imaginary part $b$ from the result. $\newline$ Calculation: From $23 - 11i$ , $a = 23$ and $b = -11$ .
Calculate Value: Step $4$ : Calculate the value of $(2a + 3b)$ . $\newline$ Calculation: $2a + 3b = 2(23) + 3(-11) = 46 - 33 = 13$ .

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