How do you simplify #(2-sqrt2i)/(3+sqrt6i)#?

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Answer 1 · 2016-12-27T10:50:52

The answer is #=(6-sqrt12)/15-(2sqrt6+3sqrt2)/15i#

Let a complex number be #z=z_1/z_2#

To simplify, we multiply the numerator and the denominator by the conjugate of the denominator

#z=(z_1*barz_2)/(z_2*barz_2)#

If #w=a+ib#, the conjugate is #barw=a-ib#

#i^2=-1#

Here,

#z=(2-sqrt2i)/(3+sqrt6i)#

#z=((2-sqrt2i)(3-sqrt6i))/((3+sqrt6i)(3-sqrt6i))#

#=(6-2sqrt6i-3sqrt2i-sqrt12)/(9+6)#

#=(6-sqrt12)/15-(2sqrt6+3sqrt2)/15i#