Proving math is broken with Euler's Formula

The complex logarithm is multi-valued, since $e^z = e^{z + 2n\pi i}$ for any integer $n$ . Assuming equality in $\boxed{\text{Step 2}}$ is therefore unwarranted.

It is customary to define $\ln z = a + bi$ in such a way that $-\pi < b \leq \pi$ ; with this definition it is true that $\ln 1 = 0$ . However, by sustituting $x = 2\pi$ , Derya essentially concludes that $e^0 = e^{2\pi i} \ \ \ \therefore\ \ \ \ 0 = 2\pi i,$ which is invalid.

The correct derivation would be:

$\begin{aligned} \text{Step 1:}&& { e }^{ ix }&=\cos { (x) } +i\sin { (x) } \\\\ \text{Step 2:}&& ix&=\ln { \big(\cos { (x) } +i\sin { (x) } \big) } + 2\pi i k \ \ \ \text{for some integer}\ k \\\\ \text{Step 3:}&& i&=\frac { \ln { \big(\cos { (x) } +i\sin { (x) } \big) } + 2\pi i k }{ x }\ \ \ \text{or}\ x = 0, k = 0 \\\\ \text{Step 4:}&& i&=\frac { \ln { \big(\cos { (2\pi ) } +i\sin { (2\pi ) } \big) } + 2\pi i k }{ 2\pi } \\\\ \text{Step 5:}&& i&=\frac { \ln { (1) } + 2\pi i k }{ 2\pi } \\\\ \text{Step 6:}&& i&=\frac { 0 + 2\pi i k }{ 2\pi } \\\\ \text{Step 7:}&& i&=i k. \end{aligned}$