Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I have the follow multiplication:

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

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;

share|improve this question

closed as unclear what you're asking by Thursday, Adam Hughes, Sami Ben Romdhane, martini, azimut Aug 2 at 17:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

3  
Why did you replace $-2$ and $-3$ with $2i$ and $3i$? –  anon Mar 29 '12 at 0:17
4  
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

2 Answers 2

up vote 3 down vote accepted

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

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

share|improve this answer
    
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 $6 \sqrt {6i^2}=6\sqrt {-6}=\pm 6i\sqrt 6$

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.