How to multiply out $(\sqrt{2}) (\sqrt{2}i)(\sqrt{2}+\sqrt{2}i)$?
$i =$ the complex imaginary number.
How to multiply out $(\sqrt{2}) (\sqrt{2}i)(\sqrt{2}+\sqrt{2}i)$?
$i =$ the complex imaginary number.
Just multiply it out according to the usual rules of algebra:
$\begin{align*} \sqrt2\cdot\sqrt2 i\cdot\left(\sqrt2+\sqrt2 i\right)&=\sqrt2\cdot\sqrt2 i\cdot\sqrt2+\sqrt2\cdot\sqrt2 i\cdot\sqrt2 i\\ &=\left(\sqrt2\cdot\sqrt2\cdot\sqrt2\right)i+\left(\sqrt2\cdot\sqrt2\cdot\sqrt2\right)(i\cdot i)\\ &=\ldots\;? \end{align*}$
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: $2 \sqrt{2} i - 2 \sqrt{2}$
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: $2\sqrt{2}i - 2\sqrt{2}$