2

We know that,

$(a^m)^n = a^{mn} = (a^{n})^m$

So ,

  1. $\sqrt {i^4} = (i^{1/2})^4 = \left(\pm\frac{1+ i}{\sqrt2} \right)^4 = -1$

  2. $\sqrt {i^4} = (i^{2}) = -1$

  3. $\sqrt {i^4} = \sqrt1 = 1$

I think only no $3$ is right. I must have violated some rule in $1$ & $2$. Please let me know.

Maybe $(a^m)^n = a^{mn} = (a^{n})^m$ is true only for Real Numbers.

lioness99a
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Utshaw
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  • You are correct. Only 3 is valid and the rule is not true for complex numbers. – windircurse Jul 03 '17 at 07:31
  • But when $m$ and $n$ are both integers, the rule is okay. That's obvious, but worth saying anyhow. – Harald Hanche-Olsen Jul 03 '17 at 07:32
  • For much more information on the square root of complex numbers, look at this question. – Harald Hanche-Olsen Jul 03 '17 at 07:35
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    Start from here. Over the course of the history of our site several users have failed to appreciate the intricacies the complex exponentiation is multivalues unless the exponent is an integer. The rule $(a^m)^n=a^{mn}$ holds only if $m,n$ are both integers, or (with the usual choice of single values real powers) $a,m,n$ are all real and $a>0$. – Jyrki Lahtonen Jul 03 '17 at 07:49

3 Answers3

2

I would say that $\sqrt{x}$ invokes a choice. It is usually defined as the real number $y\ge0$ such that $y^2 = x$.

So this definition does not carry over to the complex numbers since there is no notion of greater zero.

For general $n\in \Bbb N$ $a^{\frac{1}{n}}$ has $n$ different solutions if you defined it by $\left(a^{\frac{1}{n}}\right)^n= a$, if $n$ is odd and $a$ is real they all agree.

So what happens here could be see as taking different choices.

But as there is no "canonical choice" for $\sqrt{x}$ when $x$ is complex it is usually not defined. What I mean by "canonical choice" for reals it is enough to require $\sqrt{1}=1$ and continuity, but for complex numbers such a function does not exist on the whole complex plane.

For exponents which are not rational this is quite similar: you can define $x^y$ as $e^{y\log x}$ which in the complex case also invokes a choice since $\exp$ is $2\pi$ periodic, i.e. you have to choose a branch of the logarithm. (which is not defined on the whole complex plane)

Back to your question. How does this relate to the rule $$(a^m)^n = a^{mn} ?$$ In the real case you can make unique choices for both expressions. I will express it with the definition $x^y =e^{y\log x}$. So we have $$ (a^m)^n = e^{n(log(e^{m\log n}))} = e^{n({m\log n})} =a^{mn} $$ where I just used that $\log \exp x = x$ and for real numbers there is a unique choice for the logarithm. When you want to check that $x^y =e^{y\log x}$ actually makes sense for rational numbers you will see that there is also a choice invoked by defining roots to be positive. But for complex numbers you don't have such a function $\log$.

So for 1 and 2 the rule does not apply and 3 is correct.

user60589
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1

In $\mathbb C$ we have $\sqrt1=\{1,-1\}$.

Thus, $\sqrt{i^4}=\sqrt1=\{1,-1\}$.

Michael Rozenberg
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1

$$\sqrt{i^4}=i^2 = -1\tag1$$
Or $$\sqrt{i^4}= \sqrt{(\sqrt{-1})^4} = (\sqrt{-1})^2 = -1 = i^2 \tag2$$
Or $$\sqrt{i^4} = \sqrt{i^2\cdot i^2} = \sqrt{i^2}\cdot \sqrt{i^2}= i\cdot i = i^2 = -1\tag3$$

Darío A. Gutiérrez
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