Question

Question #db9f4

Answer 1

`2/(-sqrt(2)i) = sqrt(2)i`

`2/(sqrt(2)i) = -sqrt(2)i`

Explanation:

We will use the properties that `sqrt(2)*sqrt(2) = 2` and `i*i = -1` to eliminate the imaginary and irrational components from the denominators:

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`y = 2/(-sqrt(2)i)`

`=(2*sqrt(2)i)/((-sqrt(2)i*sqrt(2)i))`

`=(2sqrt(2)i)/(-(sqrt(2)*sqrt(2))(i*i))`

`=(2sqrt(2)i)/(-2(-1))`

`=(2sqrt(2)i)/2`

`=sqrt(2)i`

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For `y=2/(sqrt(2)i)`, we could go through the same process, however we can also just multiply the first result by `-1`:

`y = 2/(sqrt(2)i)=-(2/(-sqrt(2)i))=-sqrt(2)i`

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Note that this is a simple case of more general techniques for eliminating square roots or imaginary components from a denominator. The method is the similar for both, and relies on the identity `(a+b)(a-b)=a^2-b^2`.

`1/(a+bsqrt(r))=(a-bsqrt(r))/((a+bsqrt(r))(a-bsqrt(r)))`

`=(a-bsqrt(r))/(a^2-(bsqrt(r))^2)`

`=(a-bsqrt(r))/(a^2-b^2r)`

`1/(a+bi) = (a-bi)/((a+bi)(a-bi))`

`=(a-bi)/(a^2-(bi)^2)`

`=(a-bi)/(a^2+b^2)`

We say that `x-y` is the conjugate of `x+y`, and `a-bi` is the complex conjugate of `a+bi`.