How do you simplify #(6-isqrt2)/(6+isqrt2)# and write the complex number in standard form?

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Answer 1 · 2016-11-06T18:39:40

Please see the explanation for steps leading to the answer: #17/19 - (6/19sqrt(2))i #

Multiply the numerator and denominator by the complex conjugate of the denominator, #(6 - isqrt(2))#

#(6 - isqrt(2))/(6 + isqrt(2))(6 - isqrt(2))/(6 - isqrt(2)) = #

This done, because it is well known that this will turn the denominator into a real number:

# (6 - isqrt(2))^2/(36 - 2i^2) =#

Substitute -1 for #i^2#:

#(6 - isqrt(2))^2/(36 + 2) = #

#(6 - isqrt(2))^2/(38) = #

Use the pattern #(a - b)^2 = a^2 - 2ab + b^2# to multiply the numerator:

#(36 - (12sqrt(2))i + 2i^2)/38 = #

Substitute -1 for #i^2#:

#(36 - (12sqrt(2))i - 2)/38 = #

#(34 - (12sqrt(2))i)/38 = #

#17/19 - (6/19sqrt(2))i #