"Proof" that $i=-\sqrt3/3$

Question

I encountered the following "proof" and I simply can't find the error in the reasoning.

$$ e^{\frac{i\pi}{2}}=\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}} = i $$ $$ e^{\frac{i\pi}{6}}=\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}+\frac{i}{2}=(e^{\frac{i\pi}{2}})^{\frac{1}{3}}=i^{\frac{1}{3}} $$ $$ 2i^{\frac{1}{3}}-i=\sqrt{3} $$ $$ i(2i^{-\frac{2}{3}}-1)=\sqrt{3} $$ $$ i(\frac{2}{(i^2)^{\frac{1}{3}}}-1)=\sqrt{3} $$ $$ i=-\frac{\sqrt{3}}{3} $$ I'm guessing the error is somewhere in the process of exponentation, but I can't find it.

@StellaBiderman probably twice the second line minus the first line — ziggurism, Dec 09 '17 at 00:34
You're expecting that $i^{1/3}$ is the same as $i\cdot i^{-2/3}$. I don't know that that's the error, because I haven't looked that thoroughly into your calculations, but it is a common error, and I would be very careful with exactly what cube roots I picked if I were to write that. — Arthur, Dec 09 '17 at 00:34
$(-1)^{1/3} = \frac12 + i\frac{\sqrt 3}2$ using the principal branch. — Trevor Gunn, Dec 09 '17 at 00:35
As always with many "proofs", we are applying a non-injective function, then unapplying it in an odd way, while ignoring all possible solutions. — Kaynex, Dec 09 '17 at 00:37
$i^{-\frac 23}\ne \frac 1 {(i^2)^{\frac 12}} $. $(a^x)^y\ne a^{xy} $ if $a $ is not real. (Doesn't even hold if if a is negative real). — fleablood, Dec 09 '17 at 00:40

ziggurism · Answer 1 · 2022-12-15T03:47:05.223

4

In general you should not use the laws $z^{a+b}=z^az^b$ and $(z^a)^b=z^{ab}$ if $a$ and $b$ are not integers or the base $z$ is not a nonnegative real.

If you insist on using these rules, you must account for the branches of the exponential function.

If you rewrite your proof, carefully observing these rules, you should end up with a true equation.

edited Dec 15 '22 at 03:47

answered Dec 09 '17 at 00:36

ziggurism

16,756

score 1 · Accepted Answer · answered Dec 09 '17 at 00:54

One must be careful about which branch of the complex logarithm we are using. For instance, with $(-1)^{1/3}$ what this really is is

$$ \exp\left\{\frac13 \log(-1) \right\} = \exp\left\{\frac13 (i\pi) \right\} = \frac12 + i\frac{\sqrt{3}}2. $$

It is important that we agree that $\log(-1) = i\pi$ and not, for instance, $3i\pi$ (which would give $(-1)^{1/3} = -1$).

If we agree that $\operatorname{Im}(\log z) \in (-\pi,\pi]$ then we do indeed have

$$ \exp\left\{\frac{i\pi}6 \right\} = \exp\left\{\frac{i\pi}2 \right\}^{1/3} = i^{1/3}. $$

This takes us up to the line

$$ 2i^{1/3}-i=\sqrt{3} $$

without issue.

The next question is whether

$$ i^{1/3} = i \cdot i^{-2/3}$$

We know already that $i^{1/3} = \exp\left\{\frac{i\pi}6 \right\}$ under our convention. For the right hand side we have

\begin{align} i \cdot i^{-2/3} &= i \cdot \exp\left\{-\frac{2}3 \log(i) \right\} \\ &= i \cdot \exp\left\{-\frac{2}3 \frac{i\pi}{2} \right\} \\ &= \exp\left\{\frac{i\pi}{2} \right\} \exp\left\{-\frac{i\pi}3 \right\} \\ &= \exp\left\{\frac{i\pi}{2} -\frac{i\pi}3 \right\} \\ &= \exp\left\{\frac{i\pi}{6} \right\} \end{align}

So indeed

$$ 2i^{1/3}-i = i\left( \frac{2}{i^{2/3}} - 1 \right) = \sqrt3. $$

Now another question is whether or not

$$ i^{2/3} = (i^2)^{1/3} $$

You can (and should) check that these are both equal to

$$ \frac12 + i\frac{\sqrt{3}}2. $$

Which is what we had at the beginning.

Finally, we substitute this for $i^{2/3}$ and we obtain

\begin{align} i \left( \frac{2}{\frac12 + i\frac{\sqrt{3}}2} - 1 \right) &= i \left( 2\left[ \frac12 - i\frac{\sqrt{3}}2 \right] - 1 \right) \\ &= i \left( - i\sqrt{3} \right) \\ &= \sqrt{3}. \end{align}

Summary: the only issue was $(-1)^{1/3} = \frac12 + i\frac{\sqrt{3}}2$ not $-1$.

a much better answer than mine. You should be entitled to my upvotes! — ziggurism, Dec 09 '17 at 00:59

"Proof" that $i=-\sqrt3/3$

2 Answers2