Question #db9f4

Question

Question #db9f4

1 Answer

Impact of this question

1508 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-23T17:54:11

$\frac{2}{- \sqrt{2} i} = \sqrt{2} i$ $2/(-sqrt(2)i) = sqrt(2)i$

$\frac{2}{\sqrt{2} i} = - \sqrt{2} i$ $2/(sqrt(2)i) = -sqrt(2)i$

Explanation:

We will use the properties that $\sqrt{2} \cdot \sqrt{2} = 2$ $sqrt(2)*sqrt(2) = 2$ and $i \cdot i = - 1$ $i*i = -1$ to eliminate the imaginary and irrational components from the denominators:

$y = \frac{2}{- \sqrt{2} i}$ $y = 2/(-sqrt(2)i)$

$= \frac{2 \cdot \sqrt{2} i}{(- \sqrt{2} i \cdot \sqrt{2} i)}$ $=(2*sqrt(2)i)/((-sqrt(2)i*sqrt(2)i))$

$= \frac{2 \sqrt{2} i}{- (\sqrt{2} \cdot \sqrt{2}) (i \cdot i)}$ $=(2sqrt(2)i)/(-(sqrt(2)*sqrt(2))(i*i))$

$= \frac{2 \sqrt{2} i}{- 2 (- 1)}$ $=(2sqrt(2)i)/(-2(-1))$

$= \frac{2 \sqrt{2} i}{2}$ $=(2sqrt(2)i)/2$

$= \sqrt{2} i$ $=sqrt(2)i$

For $y = \frac{2}{\sqrt{2} i}$ $y=2/(sqrt(2)i)$ , we could go through the same process, however we can also just multiply the first result by $- 1$ $-1$ :

$y = \frac{2}{\sqrt{2} i} = - (\frac{2}{- \sqrt{2} i}) = - \sqrt{2} i$ $y = 2/(sqrt(2)i)=-(2/(-sqrt(2)i))=-sqrt(2)i$

Note that this is a simple case of more general techniques for eliminating square roots or imaginary components from a denominator. The method is the similar for both, and relies on the identity $(a + b) (a - b) = a^{2} - b^{2}$ $(a+b)(a-b)=a^2-b^2$ .

$\frac{1}{a + b \sqrt{r}} = \frac{a - b \sqrt{r}}{(a + b \sqrt{r}) (a - b \sqrt{r})}$ $1/(a+bsqrt(r))=(a-bsqrt(r))/((a+bsqrt(r))(a-bsqrt(r)))$

$= \frac{a - b \sqrt{r}}{a^{2} - {(b \sqrt{r})}^{2}}$ $=(a-bsqrt(r))/(a^2-(bsqrt(r))^2)$

$= \frac{a - b \sqrt{r}}{a^{2} - b^{2} r}$ $=(a-bsqrt(r))/(a^2-b^2r)$

$\frac{1}{a + b i} = \frac{a - b i}{(a + b i) (a - b i)}$ $1/(a+bi) = (a-bi)/((a+bi)(a-bi))$

$= \frac{a - b i}{a^{2} - {(b i)}^{2}}$ $=(a-bi)/(a^2-(bi)^2)$

$= \frac{a - b i}{a^{2} + b^{2}}$ $=(a-bi)/(a^2+b^2)$

We say that $x - y$ $x-y$ is the conjugate of $x + y$ $x+y$ , and $a - b i$ $a-bi$ is the complex conjugate of $a + b i$ $a+bi$ .

Science

Math

Humanities

... and beyond

Question #db9f4

1 Answer

Explanation:

Related questions

Impact of this question