Question #db9f4
1 Answer
Explanation:
We will use the properties that
#=(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
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-bsqrt(r))/(a^2-(bsqrt(r))^2)#
#=(a-bsqrt(r))/(a^2-b^2r)#
#=(a-bi)/(a^2-(bi)^2)#
#=(a-bi)/(a^2+b^2)#
We say that