False proof of $\pi=0$ In an attempt to prove to myself that the equation $e^{i\pi} = -1$ is indeed true, I attempted to 'reverse engineer' it into a state that I already know to be true using Year 12 methods, (i.e. able to verify it in a basic calculator), but ended up with $\pi = 0$ and was wondering where I went wrong. I don't doubt it comes down to me knowing pretty much nothing about imaginary numbers other then '$i$' is defined as the square root of $-1$.

Anyway here was my attempt:

\begin{align}
  e^{i\pi} &amp;= -1 \\
  \ln(-1) &amp;= i\pi \\
  2 \ln(-1) &amp;= 2i \pi, \text{ multiplied all by $2$} \\
  \ln((-1)^2) &amp;= 2 \sqrt{-1} \pi \\
  \ln(1) &amp;= 2 \sqrt{-1} \pi, \text{ it was at this point I realised I'd messed up} \\
  0 &amp;= 4 \cdot -1 \cdot \pi^2, \text{ pointless squaring} \\
  0 &amp;= -4 \cdot \pi^2 \\
  0 &amp;= \pi
\end{align}
Here's a quick way to summarize the essence of your paradox:
$$
e^{2 \pi i} = e^0 \overset{?}\implies 2 \pi i = 0
$$
The first equation is true, the second is not. The problem is that the log is no longer uniquely defined.  If it were defined in that sense, then taking the "log of both sides" would take us from the first equation to the second.

With real numbers, we can only have $e^x = e^y$ when $x = y$. This is not true if $x$ and $y$ are complex.



Since you are learning about complex numbers for the first time, it is worth noting that for real numbers $x$, we have Eulers formula:
$$
e^{xi} = \cos(x) + i \sin (x)
$$
Notably, both cosine and sine are periodic, so this function will be periodic (over the real values of $x$) with period $2 \pi$.  For more on all this, I recommend this video (mathologer) and this video (3blue1brown).
You should be knowing that e^(npii) where n is any integer is equal to one,so that means ln(1) has multiple values ie,npii,not only 0
So you can only take the case in which it is equal,ie ln 1 equals to 2*pi*i