3
$\begingroup$

I just thinking what $i^{i}$ should be, arrived at a quiet awkward thing.
So this was that awkward thinking :
Let $ i^{i} = a$
$(i^{i})^{2i} = a^{2i}$
$i^{-2}=a^{2i}$
$-1=a^{2i}$
Now if we take power of $\frac{1}{2i}$ both sides
$(-1)^{\frac{1}{2i}}=a$
$((-1)^{\frac{1}{2}}) ^{\frac{1}{i}}=a$
$i^{\frac{1}{i}} =a$
$i^{\frac{1}{i}} =i^{i}$
Now if we compare the power (since the base is same)
$\frac 1i = i$
$i^2=1$
So, $-1=i^2=1$.
I am quite new to imaginary numbers, So please show my mistake elaborately.
P. S. Thanks in advance.

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    i think the power rules are not true in $\mathbb{C}$ – 2017-01-21

2 Answers 2

10

There are two illegal steps you've made:

  1. When you use $x^{ab} = (x^a)^b.$ This type of arithmetic is also not generally valid for complex numbers. A counterexample is when $a = 2$ and $b=1/2$ and $x=-i.$ Then $(-i)^{2\cdot 1/2} = (-i)^1=-i,$ but $((-i)^2)^{1/2} = (-1)^{1/2} = i.$ Likewise, if $x^a = y^a$ we cannot take both sides to the $1/a$ power to give $x=y.$ For instance $(-2)^2 = (2)^2$ but $-2\ne 2.$

  2. When you conclude the exponents are equal cause the bases are equal, i.e. conclude from $a^b = a^c$ that $b=c.$ A counterexample to this is that $e^{x+2\pi i} = e^x$ for any $x$ but the exponents are not equal.

The difficulty for 1 is that roots are generally not unique, so when we implicitly pick a 'principal' root during arithmetic it might not be the right one. For the second it's that the exponential function is not one-to-one when the domain is extended to the complex numbers, so we can't invert it by taking logs like the real case. If $e^a = e^b,$ we know $a$ and $b$ are different by an integer multiple of $2\pi i,$ not that they are equal.

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    Thanks for ur answer. I didnt know the second point earlier and the first one was my mistake even after knowing. – 2017-01-21
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    Why is it not generally valid(the third point) ? I mean it's a basic arithmetic operation which should be valid everywhere? – 2017-01-21
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    @geekyme look at my counterexample. See what happened? If we'd interpreted $(-1)^{1/2} = -i$ rather than as $i$ in the last equality we'd be fine. And why shouldn't we be able to? $-i$ is a square root of $(-1).$ The problem is that we are forced by the arbitrary 'take the principal root' rule for how we define $x^{1/2}$ to choose one root when the other could be the one that makes the identity hold. – 2017-01-21
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    Your (1) is wrong -- if you have two expressions that you know are equal, then applying the same function (such as raising to a particular power) on both sides does produce a valid equality. In the example you give, it is indeed true that $(2^2)^{1/2} = ((-2)^2)^{1/2}$; error is only introduced when you assume that $(a^2)^{1/2}$ is necessarily the same as $a$. – 2017-01-21
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    @HenningMakholm Yes, you're right. And that could then be subsumed into item 3. Corrected. – 2017-01-21
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    @geekyme My point 1 was wrong(see Henning's comment.), so I edited. The error I expressed there was actually an example of item 3. – 2017-01-21
1

I'll give you another one.

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

For positive reals $a$ and $b$ we have that $\sqrt{ab}=\sqrt{a}\sqrt{b}$. But who says this still holds for other numbers?

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    You're addressing [this question](http://math.stackexchange.com/questions/1966317/backward-proof-of-i2-1?rq=1). – 2017-01-21
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    Well yeah, I was only trying to point out that one should take care when consider powers of non-positive numbers. – 2017-01-21
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    Its a common troll theme: http://hsm.stackexchange.com/questions/5522/when-and-who-debunked-first-the-complex-numbers-in-mathematics-rigorously – 2017-01-21
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    @j4nbur53: Holy cow, even for a crazy guy he's insane. They never listen to good arguments though, impossible to change their mind. – 2017-01-21