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This question already has an answer here:

From what I know, the rule to distribute exponents is like:

$$(a b)^x = a^x b^x$$

Thus, if $a = \sqrt 2$ and $b = \sqrt 3$, $ab = \sqrt 6$.

However, the imaginary unit $i = \sqrt{-1}$ has a different behavior, because if I take $a = -2$, $b = -3$ and $x = 1/2$:

$${[(-2) (-3)]}^{1/2} = (-2)^{1/2} (-3)^{1/2} = \sqrt{-2} \sqrt{-3} = i\sqrt 2 i \sqrt 3 = - \sqrt 6$$

Although, before I learned complex numbers, I thought,

$${[(-2) (-3)]}^{1/2} = [6]^{1/2} = \sqrt 6$$

What am I making wrong here and which is the right answer?

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marked as duplicate by user21820, Did, Xander Henderson, Namaste, Community May 5 '18 at 3:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ good catch! It does indeed not hold for negative reals $\endgroup$ – qbert May 4 '18 at 0:51
  • $\begingroup$ Check out math.stackexchange.com/questions/2761058 for some discussion of the issue. $\endgroup$ – Cedron Dawg May 4 '18 at 0:53
  • $\begingroup$ Really, how do I work in this situation then? I am not from the math area, just being curious because I had a test that asked $i \sqrt{-1}$ and I found two possible answers and couldn't understand which one was correct. $\endgroup$ – Pedro Henrique Vaz Valois May 4 '18 at 0:56
  • $\begingroup$ See What are the Laws of Rational Exponents? $\endgroup$ – dxiv May 4 '18 at 0:56
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    $\begingroup$ To ask what $i\sqrt{-1}$ would be is a rather poorly phrased question. A better way would be to ask the solutions to $z^2=1$.In fact, I never liked the convention $i=\sqrt{-1}$ to begin with. A better option would be to define $i$ as the imaginary unit with the property $i^2=-1$. And even that can ultimately be improved $\endgroup$ – imranfat May 4 '18 at 0:59
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The definition of $a^x$ in complex analysis is $\exp(x \log(a))$, where $\log$ is the multi-valued complex (natural) logarithm. Thus if $\text{Log}(a)$ is one value of this logarithm, the others are $\text{Log}(a) + 2 \pi i n$ for arbitrary integers $n$.

Now the identity $(ab)^x = a^x b^x$ is not necessarily true. Instead $$ (ab)^x = a^x b^x \exp(2 \pi i n x)$$

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  • $\begingroup$ Does this mean the answer to the following question is wrong? math.stackexchange.com/questions/2616891/… $\endgroup$ – Kagaratsch May 4 '18 at 1:58
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    $\begingroup$ Note that the question you link to has the assumption $a, b > 0$. Under this assumption the "principal branch" has $\log(a)$ and $\log(b)$ real, and then $(ab)^x = a^x b^x$ is true for the principal branch. $\endgroup$ – Robert Israel May 4 '18 at 4:28
  • $\begingroup$ I see, thank you for the clarification! $\endgroup$ – Kagaratsch May 4 '18 at 5:34
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Here is another version of your argument, with some answers.

Simply, such a rule does not apply to complex numbers in general.

Regarding your test question, writing $i\sqrt{-1}$ makes little sense. When we write $\sqrt2$, we are using the convention "the positive number such that its square is $2$". When you want square roots of $-1$, there is no obvious way from this point of view to distinguish between $i$ and $-i$. What we usually do to avoid this problem is consider $i$ as a number with $i^2=-1$.

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When you move into the complex arena, non-integer exponents can produce multiple answers.

The errors and contradictions come in when you assume that the answer is the principal value. When you constrain yourself to positive reals raised to a real value, the principal answer is the only answer and thus the correct answer. This makes the rules safe to use.

In the case of the $i\sqrt{-1}$ question is ambiguous because you can't tell whether the square root means the principal value (which is usually the convention) or if it should have a $\pm$ in front of it.

If you are aware of the multiple possibilities (by including a factor of $e^{i2\pi}$ in your base(s)), or can somehow qualify the range of your answers, the rules can be used.

Ced

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  • $\begingroup$ @Arnaud Mortier, Thanks for the fix. $\endgroup$ – Cedron Dawg May 4 '18 at 1:19
  • $\begingroup$ You're welcome! $\endgroup$ – Arnaud Mortier May 4 '18 at 1:22

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