How do you simplify $(5 i) (- 8 i)$ $(5i)(-8i)$ ?

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Answer 1 · 2018-05-14T22:19:39

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$(5 i) \cdot (- 8 i) = 40$ $color(blue)((5i)*(-8i)=40$

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A Complex Number is a combination of a Real Number and an Imaginary Number.

Standard Form of a Complex Number is : $a + b i$ $color(red)(a+bi$ , where

$a$ $color(red)(a$ is the Real Part and $b$ $color(red)(b$ is the Imaginary Part.

In the problem given, the Real Part is unavailable.

You have just the Imaginary Part for both the complex numbers.

Multiply the complex numbers $(5 i) \cdot (- 8 i)$ $color(blue)((5i)*(-8i)$ :

$(5 i) \cdot (- 8 i)$ $(5i)*(-8i)$

Multiply the numerical constants first with the sign:

$5 \cdot (- 8)$ $5*(-8)$

$\Rightarrow (- 40)$ $rArr color(red)((-40)$

You need this intermediate result in the final step.

Next, multiply $i$ $i$ by $i$ $i$ .

In complex numbers, $i = \sqrt{- 1}$ $color(blue)(i = sqrt(-1)$

Hence,

$i \cdot i = \sqrt{- 1} \cdot \sqrt{- 1}$ $i*i = sqrt(-1)*sqrt(-1)$

$\Rightarrow i^{2} = {[\sqrt{- 1}]}^{2}$ $rArr i^2=[sqrt(-1)]^2$

In radicals arithmetic, $\sqrt{a^{2}} = \sqrt{a \cdot a} = a$ $color(red)(sqrt(a^2)=sqrt(a*a)=a$

$\Rightarrow i^{2} = (- 1)$ $rArr i^2=(-1)$

Hence,

$(5 i) (- 8 i) = [- 40 \cdot (- 1)] = 40$ $color(blue)((5i)(-8i)=[-40*(-1)]=40$

Hope it helps.

How do you simplify (5i)(-8i)(5i)(−8i)(5i)(-8i)?