Question #db9f4

1 Answer
sente
Aug 23, 2016

2/(-sqrt(2)i) = sqrt(2)i2−√2i=√2i

2/(sqrt(2)i) = -sqrt(2)i2√2i=−√2i

Explanation:

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

y = 2/(-sqrt(2)i)y=2−√2i

=(2*sqrt(2)i)/((-sqrt(2)i*sqrt(2)i))=2⋅√2i(−√2i⋅√2i)

=(2sqrt(2)i)/(-(sqrt(2)*sqrt(2))(i*i))=2√2i−(√2⋅√2)(i⋅i)

=(2sqrt(2)i)/(-2(-1))=2√2i−2(−1)

=(2sqrt(2)i)/2=2√2i2

=sqrt(2)i=√2i

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

y = 2/(sqrt(2)i)=-(2/(-sqrt(2)i))=-sqrt(2)iy=2√2i=−(2−√2i)=−√2i

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(a+b)(a−b)=a2−b2.

1/(a+bsqrt(r))=(a-bsqrt(r))/((a+bsqrt(r))(a-bsqrt(r)))1a+b√r=a−b√r(a+b√r)(a−b√r)

=(a-bsqrt(r))/(a^2-(bsqrt(r))^2)=a−b√ra2−(b√r)2

=(a-bsqrt(r))/(a^2-b^2r)=a−b√ra2−b2r

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

=(a-bi)/(a^2-(bi)^2)=a−bia2−(bi)2

=(a-bi)/(a^2+b^2)=a−bia2+b2

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

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