Question #db9f4

1 Answer
sente
Aug 23, 2016

#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:

#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#

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#

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#.

Related questions
See all questions in Common Logs
Impact of this question
1100 views around the world
You can reuse this answer
Creative Commons License