Simplify the complex expression (x-i)(3+i)(i-1)/(1+i)(3-i)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The expression `((x-i)(3+i)(i-1))/((1+i)(3-i))` has to be simplified.

`((x-i)(3+i)(i-1))/((1+i)(3-i))`

= `((x-i)(3+i)(i-1)(1 - i)(3+i))/((1+i)(3-i)(1-i)(3+i))`

= `((x-i)(3+i)(i-1)(1 - i)(3+i))/((1 + 1)(9 + 1))`

= `((x-i)(3+i)(i-1)(1 - i)(3+i))/20`

= `(-(x-i)(3+i)^2(i-1)^2)/20`

= `(-(x-i)(9 - 1 + 6i)(-1 + 1 - 2i))/20`

= `(-(x-i)(8 + 6i)(- 2i))/20`

= `((x-i)(16i + 12i^2))/20`

= `((x-i)(16i - 12))/20`

= `((x-i)(4i - 3))/5`

= `(4x*i - 4i^2 - 3x + 3i)/5`

= `(4x*i + 4 - 3x + 3i)/5`

The simplified form of `((x-i)(3+i)(i-1))/((1+i)(3-i)) = (4x*i + 4 - 3x + 3i)/5`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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To simplify `((x-i)(3+i)(i-1))/((1+i)(3-i))` use complex conjugates of terms in the denominator. Multiply the numerator and denominator by `(1 - i)(3+i)` .

`((x-i)(3+i)(i-1)(1 - i)(3+i))/((1 - i)(1+i)(3-i)(3+i))`

= `((x-i)(3+i)(i-1)(1 - i)(3+i))/((1 +1)(9+1))`

The denominator is now 2*10 = 20

Simplifying the numerator gives: `-(x-i)(3+i)^2(1 - i)^2`

= `-(x - i)(9+6i - 1)(1 - 2i - 1)`

= `-(x - i)(8+6i)(-2i)`

= `(2x*i - 2i^2)(8+6i)`

= `(2x*i + 2)(8+6i)`

= `16x*i + 16 + 12xi^2 + 12i`

= `16x*i + 16 - 12x + 12i`

Dividing this by 20 and cancelling common factors gives:

`(16x*i + 16 - 12x + 12i)/20`

= `(4x*i + 4 - 3x + 3i)/5`

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