Rummaging through the treasury of mathematical thought, one might find this:

**Picard’s Little Theorem**: if is a non-constant complex function which is analytic in the entire complex plane (entire), then it either takes every complex value or misses at most one value.

(Charles Émile Picard (1856–1941) also has the Great Picard Theorem to his credit as well as everyone’s favourite Picard–Lindelöf theorem.)

So the LPT sets up a dichotomy: either an entire function takes every possible complex value, as exemplified by , or it misses out a value: never takes the value zero.

But this gives rise to a slightly puerile conundrum you might want to think about. Consider . This is clearly an entire function (check). Since this is an exponential of something, it never takes the value zero. On the other hand, since never takes the value zero, does not take the value . So it looks as if misses out *two* values, and . Hmm, that would disprove Picard’s Little Theorem! Please work out what is going on here. [Hint: just as with the missing pound, nothing but snake-oil salesmanship…]

(MG)

### Like this:

Like Loading...

*Related*

I assume there’s another COMPLEX value z such that e^z = 1, but I don’t know what it is.

Yep — you’ve spotted the hole in the logic there! As to finding values of z such that , it may be helpful to start by thinking of Euler’s formula and see what this suggests…

Okay. e^(2*pi*i)=1. (Also any other even number in place of 2.) Then to solve f(z)=1 you need to solve e^z=2*pi*i. The imaginary part of z needs to be pi/2 (or 3*pi/2, etc.) to make the real part of e^z zero. e^(i*pi/2)=i. So now we need to find the real part x of z to make e^x=2*pi. Taking ln of both sides gives x=ln(2*pi). So f(z)=1 when z=ln(2pi)+i(pi/2), among others. Right?

Right. Note that we could have demonstrated the hole in the argument even if we couldn’t identify z such that : in this case we could just have used Picard’s Little Theorem. applied to : since this is an entire function and we know it’s never zero, we know that there must be some z such that . (Neat, huh?)

Pingback: Solution to the problem with Picard | Degree of Freedom