# Another question on radicals [closed]

I'm back, and again with radicals and complex numbers;

$$2\sqrt{-2} \cdot 3\sqrt{-3} =$$

Working:

$$2\sqrt{2i} \cdot 3\sqrt{3i} =$$

And now, the doubt:

$$6\sqrt{6i^2}$$

even with the $i^2$ inside the square and $i^2 = -1$, I can do:

$$-6\sqrt{6}$$

? Sorry for the silly question;

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## closed as unclear what you're asking by pizza, Adam Hughes, Sami Ben Romdhane, martini, azimutAug 2 '14 at 17:57

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Why did you replace $-2$ and $-3$ with $2i$ and $3i$? –  anon Mar 29 '12 at 0:17
How did $\sqrt{-2}$ become $\sqrt{2i}$? Not that it matters, since calculations of this type get their "contradictions" by forgetting that $\sqrt$ is not single-valued. –  André Nicolas Mar 29 '12 at 0:21

You replaced negative signs ($-$) with the imaginary unit $i$ inside the square roots. This isn't valid; the negative signs indicate multiplication by $-1$, while we know $-1\ne i$ so you change the values at hand here. Instead, you need to evaluate the two square roots to the usual $a+bi$ form ($a$ zero):

$$\sqrt{-2}=\sqrt{2\cdot(-1)}=\sqrt{2}\cdot i \\ \sqrt{-3}=\sqrt{3\cdot(-1)}=\sqrt{3}\cdot i.$$

It is important to note that above we used the formula $\sqrt{a}\sqrt{b}=\sqrt{ab}$: this formula does not hold in general, but it does hold when at least one of $a$ or $b$ is a nonnegative real number. The reason is that the square root function, with the typical choice of branch, is such that $\arg \sqrt{z}\in[0,\pi)$ (so that the function never returns a value below the $x$-axis in the complex plane, and never returns a negative real). If you're not careful with these things, you could get a contradiction like

$$-1=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1.$$

Other choices of branch for the square root function are also possible.

Now if we use the above, we obtain

$$2\sqrt{2}\,i\cdot3\sqrt{3}\,i=i^2\cdot 2\cdot3\cdot\sqrt{2\cdot3}\cdot=-6\sqrt{6}.$$

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Yes, I was mistakenly interpreting the imaginary numbers, I was still thinking the $i$ belonged inside the root, sorry for the late to reply, I lost my internet connection yesterday night; Thanks for the answers; –  aajjbb Mar 29 '12 at 12:46

When you go to the line under Working, the $i$ should not be under the radical sign. Also in the complex numbers there is no canonical choice of which square root to use, so $\sqrt {-2}=\pm i\sqrt 2$. If you make the same choice of sign you will in fact get $-6 \sqrt 6$. But going from $6 \sqrt {6i^2}$ to $-6 \sqrt 6$ is again wrong. The $i^2$ becomes $-1$ under the radical, so $6 \sqrt {6i^2}=6\sqrt {-6}=\pm 6i\sqrt 6$

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