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

Nov 6, 2016

Please see the explanation for steps leading to the answer: $\frac{17}{19} - \left(\frac{6}{19} \sqrt{2}\right) i$

#### Explanation:

Multiply the numerator and denominator by the complex conjugate of the denominator, $\left(6 - i \sqrt{2}\right)$

$\frac{6 - i \sqrt{2}}{6 + i \sqrt{2}} \frac{6 - i \sqrt{2}}{6 - i \sqrt{2}} =$

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

${\left(6 - i \sqrt{2}\right)}^{2} / \left(36 - 2 {i}^{2}\right) =$

Substitute -1 for ${i}^{2}$:

${\left(6 - i \sqrt{2}\right)}^{2} / \left(36 + 2\right) =$

${\left(6 - i \sqrt{2}\right)}^{2} / \left(38\right) =$

Use the pattern ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$ to multiply the numerator:

$\frac{36 - \left(12 \sqrt{2}\right) i + 2 {i}^{2}}{38} =$

Substitute -1 for ${i}^{2}$:

$\frac{36 - \left(12 \sqrt{2}\right) i - 2}{38} =$

$\frac{34 - \left(12 \sqrt{2}\right) i}{38} =$

$\frac{17}{19} - \left(\frac{6}{19} \sqrt{2}\right) i$