# Question #db9f4

##### 1 Answer

#### 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