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How to multiply out $(\sqrt{2}) (\sqrt{2}i)(\sqrt{2}+\sqrt{2}i)$?

$i =$ the complex imaginary number.

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The regular way. – André Nicolas Dec 2 '12 at 23:16
I edited your post to $\LaTeX{}$ify it, but I should comment: there are many many complex imaginary numbers, not just one! [But I think most people here will know what $i$ means.] – Clive Newstead Dec 2 '12 at 23:33

Just multiply it out according to the usual rules of algebra:

$$\begin{align*} \sqrt2\cdot\sqrt2 i\cdot\left(\sqrt2+\sqrt2 i\right)&=\sqrt2\cdot\sqrt2 i\cdot\sqrt2+\sqrt2\cdot\sqrt2 i\cdot\sqrt2 i\\ &=\left(\sqrt2\cdot\sqrt2\cdot\sqrt2\right)i+\left(\sqrt2\cdot\sqrt2\cdot\sqrt2\right)(i\cdot i)\\ &=\ldots\;? \end{align*}$$

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For simplicity, let's start replace $i$ by $x$ so that we don't need to worry about the fact that it's imaginary for a bit.

Then we have $\sqrt{2} \cdot \sqrt{2}x \cdot (\sqrt{2} + \sqrt{2}x)$.

This simplifies to $2x \cdot (\sqrt{2} + \sqrt{2}x)$.

Which further reduces to $2\sqrt{2} x + 2 \sqrt{2}x^2$.

Now we can just plug back in $x=i$. We know that $i^2 = -1$ (by definition of $i$), so we have: $2 \sqrt{2} i - 2 \sqrt{2}$

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well, the $\sqrt{a}\sqrt{b} = \sqrt{ab}$,

so first we should expand this out. so we have:

$\sqrt{2}\sqrt{2}\sqrt{2}i + \sqrt{2}\sqrt{2}\sqrt{2}i^2$

now using the above result, we are left with:

$\sqrt{8}i + \sqrt{8}(-1)$ (as $i^2 = -1$).

you can simplify further by noting $\sqrt{8} = 2\sqrt{2}$.

hence your final solution will be: $2\sqrt{2}i - 2\sqrt{2}$

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Your answer would look a lot better if you used Mathjax. – Peter Phipps Dec 2 '12 at 23:32
Sorry, first time here... I have since realized LaTeX works on here....Thanks for the heads up. – drawnonward Dec 3 '12 at 0:21

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