Using Euler's identity

Question

Using Euler's identity, I was able to prove that $i = 0$, and I can't find where I went wrong.

Euler's identity, I recall being the following: $$e^{i \pi} = -1$$

From there, with a few changes like so: $e^{2 i \pi} = 1$

$e^{(2iπ+1)} = e$

I got:

$\ln e^{(2iπ+1)} = \ln e$

$2i \pi + 1 = 1$

$2i \pi = 0$

So, $i = 0$.

Which obviously is wrong. But I cannot find where I went wrong. Every step seems legit, so what law(s) did I break there?

Answer 1

Let's simplify the issue. It is true that $e^{2\pi i}=1$. It is equally true that $e^{0}=1$. So the powers must be equal, $2\pi i=0$ which implies $i=0$, right?

The issue with the argument is that the function $e^z$, where $z$ is a complex number, is not one-to-one, i.e. it associates multiple numbers $z$ with the same value (in this case $1$). In this case, the solutions of $e^z=1$ are $z=2n\pi i$, where $n$ is allowed to be any integer. So given $e^a=1=e^b$ doesn't require $a=b$.

Answer 2

As David already answered, I will simply try to give some further insight into what exactly goes wrong.

What does the function $e^{i\theta}$ look like? Of course, it is simply the unit circle on the complex plane.

Let’s now look at the unit circle on the normal cartesian plane, defined by the vector valued function $C(\theta)=(\cos(\theta) , \sin(\theta))$

Clearly, this equals $(1,0)$ at any integer multiple of $2\pi$.

Now, even though I said we were looking at the cartesian plane, the complex numbers are also a two-dimensional vector space over the reals (really an algebra, but that doesn’t matter for this problem). The two unit circle functions we considered here are exactly the same. They are both functions from the reals to a two-dimensional vector space over the reals, which “draw out” the unit circle on that space. They simply look different.

I hope that makes it seem slightly less mysterious.

Answer 3

Like many people said, when defined on $\Bbb{C}$, $w = \ln(z)$ is a multivalued function, since $z=e^w=e^{w+ik2\pi}$ for every $k \in \Bbb{Z}$.

From here, following steps in your previous style one could (erroneously) arrive to the conclusion that $ik2\pi = 0$, this statement could be asserted with the awareness that it is an equation "for exponents": $a=_eb \iff e^a=e^b$, where "$=_e$" represents this property.

Knowing this, we should treat every $\ln(z)$ involving equation assuming it's for exponents (moreover, the logarithm is defined as "the exponent $w$ that solve the equation $e^{\ln(z)}=e^w=z$").

Summing up, we could say by definition that $i\varphi =_e i(\varphi+k2\pi), k \in\Bbb{Z}$ and the statement $ik2\pi =_e 0$ is true, since once raised over $e$, those terms represents identity (full revolution around the origin) in single valued function, because if we are manipulating inside multivalued function, like logarithm, doing revolutions around a branch point (the origin) https://en.m.wikipedia.org/wiki/Branch_point cause the function to change value (change its branch)

Varying continuously $\varphi$ from $0$ to $k2\pi, k \in \Bbb{Z}$ in $\ln(e^{1+i\varphi})$, we can see that the result varies as expected: from $1$ to $1+ik2\pi$ (two different values) but $z$ in $\Bbb{C}$ cycled on the circumference with radius $e$ centered at origin, scoring $|k|$ revolutions, no matter if clockwise or counterclockwise (depending on $k$'s sign) returning at its starting point: $z=e$.

So, for every $k \in \Bbb{Z},\ln(e)=_e\ln(e)+ik2\pi$

$\ln(z)$ has infinite (numerable) values for every fixed $z$.

Now, answering your last question, the step I think it's not legit is

$e^{i2\pi+1}=e$

$\ln(e^{i2\pi+1})=_e\ln(e)$

You can't write this one without restrict it to exponentiation, otherwise as you wrote $\ln(e^{i2\pi+1})=\ln(e)$ you're literally saying: the exponent to which $e$ must be raised to produce $e^{i2\pi+1}$ (so $i2\pi+1$), is equal to the exponent to which $e$ must be raised to produce $e$ (so 1), that is your next identity:

$i2\pi+1=1$

this one we obviously know is incorrect (there is no need to go further) because if we write it as $z_1=z_2$ we can use all the complex arithmetic possibile to proof that they are two distinct numbers (simply, representing them as vectors on Gauss's plane, they are two different vectors), but again we can write $i2\pi+1=_e1$ and that is true, because $e^z$ is $i2\pi$ periodic in $\Bbb{C}$.

The wrong assumption you did is:

$a=b \iff \ln(a)=\ln(b)$

In $\Bbb{C}$ is:

$a=b \implies \ln(a)=\ln(b)$

And not vice versa.

Since no one uses $=_e$ notation (it's a lot ambiguous and risky), we can create a univoque $\ln(z)$ function, making restrictions on the argument of z, taking what is called the principal argument $\mathrm{Arg}(z)$:

$\Bbb{R} \ni \arg{z} = \mathrm{Arg}(z) + k2\pi$

with $\mathrm{Arg}(z) \in (-\pi,\pi], k \in \Bbb{Z}$

and define $\ln_{\Bbb{C}}(z)=\ln_{\Bbb{R}}|z|+i\mathrm{Arg}(z)$

This function now is single valued, because if $z$ goes here and there around the origin, $\mathrm{Arg}(z)$ doesn't take into account the revolutions made.

Now, let's leave $\Bbb{C}$, in order to show that the problem is indeed with equations involving multivalued functions, and not with $i$.

This time I'll show you how $\pi=0$:

$$ \left\{ \begin{array}{c} \sin(\frac{\pi}{2})=1 \\ \sin(\frac{5}{2}\pi)=1 \end{array} \right. $$

$$ \sin(\frac{\pi}{2}) = \sin(\frac{5}{2}\pi) $$

$$ \arcsin(\sin(\frac{\pi}{2})) = \arcsin(\sin(\frac{5}{2}\pi)) $$

$$ \frac{\pi}{2} = \frac{5}{2}\pi $$

$$ 0=2\pi $$

$$ \pi=0. $$

Hope you get the point.