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 $i^2$ :

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

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

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

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