# Is it possible to change units in order to simplify the value of an exponential?

I have the equation

$$F=e^{E_0 i \pi},$$

where $E_0$ is the time-independent electric field, and $F$ is just some important value I am trying to calculate. Obviously, it would be better if $F=-1$, but the $E_0$ is in the way. Therefore, is it permissible to take units where $E_0=1$ to get $F=-1$? I feel a little uncertain with this operation.

• Possible duplicate: physics.stackexchange.com/q/109995/2451 – Qmechanic Jul 13 '14 at 23:51
• It's certainly related, but I'm not sure if this is quite a duplicate. – David Z Jul 13 '14 at 23:56
• This latest title change is odd. For starters it's ambiguous (do we mean a dimensionless 1 or a 1 in this new unit system?). On top of that the previous version better described what he was asking. – ticster Jul 14 '14 at 0:00
• @ticster I'm not saying the title I added is the best possible choice (and you're free to improve it if you have a better idea), but the original title was spectacularly uninformative. Generally speaking, most titles of the form "Question about X" are fairly useless (and I haven't yet seen the one that isn't). As for the ambiguity: 1 doesn't have units, it's a number. So I don't see how this ambiguity exists. – David Z Jul 14 '14 at 2:42
• In practice it's a bit ambiguous because of the (horrible) habit of theorists to use natural units as though any "1" were unitless. – ticster Jul 14 '14 at 9:24

$E_0$ better be dimensionless as the exponential function is defined for dimensionless arguments. If it has some dimension, it needs to be multiplied or divided by a constant to fix that. Yes, you can take that constant to make the exponential $-1$, but usually $F$ has units as well and there is some constant multiplying the exponential. In that case, the units of the two constants need to be consistent. I suspect you have oversimplified the problem.
That's not how unit changing works. If $E_0 \neq 1$ in the units in which your equation is written, then to change to units in which $E'_0 = 1$ you have to introduce some (possibly dimensionful) conversion factor $k$ so that $E_0 = k E'_0$. So all you do with such a change is
$$F = \mathrm{e}^{E_0\mathrm{i}\pi} = \mathrm{e}^{kE'_0\mathrm{i}\pi} \overset{E'_0 = 1}{=} \mathrm{e}^{k\mathrm{i}\pi}$$
which is still not $-1$ if $k \neq 1$, which would only be the case for $E_0 = 1$.
You feel uncertain because what you're trying to do is dimensionaly unsound. What unit is $F$ ? There is no answer to that question. And here's how you can convince yourself of this.
The value of $F$ depends on the units of $E_0$. In one case you can have a purely imaginary $F$, in another a purely real one, and in yet another a linear combination of both (depending on the choice of unit $E_0$ could be $1/2$, $1$, or $0.5$ making $F$ $i$, $-1$, $\sqrt{2}/2 + i \sqrt{2}/2$). What kind of unit conversion would change a quantity from a pure imaginary, to a real, to a general complex number ? None. I suggest you give more detail on how you arrived at a point where you would like to compute such a quantity. In any reasonable derivation you should only end up with unitless numbers in your exponents.