How do you simplify (3+ isqrt2)/ (7-isqrt2)?

Jan 30, 2016

Multiplying denominator and numerator by the complex conjugate of the numerator, $\left(7 + i \sqrt{2}\right)$, shows that this expression can be simplified to: $\frac{21 + 10 i \sqrt{2} - 2}{51}$

Explanation:

The complex conjugate of a number $\left(a + b i\right)$ is $\left(a - b i\right)$, and multiplying top (denominator) and bottom (numerator) of a complex fraction by the complex conjugate of the bottom (numerator) will simpifly it:

$\frac{3 + i \sqrt{2}}{7 - i \sqrt{2}} \cdot \frac{7 + i \sqrt{2}}{7 + i \sqrt{2}}$

Note that any number, including a complex number, divided by itself is 1, so in multiplying by $\frac{7 + i \sqrt{2}}{7 + i \sqrt{2}}$ we are in effect multiplying by 1, leaving the result unchanged.

Note: $i \cdot i = - 1$ and $\sqrt{2} \cdot \sqrt{2} = 2$

$\frac{3 + i \sqrt{2}}{7 - i \sqrt{2}} \cdot \frac{7 + i \sqrt{2}}{7 + i \sqrt{2}}$ = $\frac{21 + 3 i \sqrt{2} + 7 i \sqrt{2} - 2}{49 + 7 i \sqrt{2} - 7 i \sqrt{2} + 2}$
= $\frac{21 + 10 i \sqrt{2} - 2}{49 + 2} = \frac{21 + 10 i \sqrt{2} - 2}{51}$