Determination of complex logarithm: $\mathrm{Log}(z) = i$ I read that the determination of complex logarythm $\mathrm{Log}(z)$ is defined as:
$$\mathrm{Log}(z) := \ln|z| + i\mathrm{Arg}(z)$$
with $\mathrm{Arg}(z)$ the main argument of the complex number. I need to solve:
$$\mathrm{Log}(z) = i$$
Assuming $z =: x + iy$, the right answer is $\cos 1 + i \sin 1$. I found:
$$\mathrm{Log}(z) = i$$
$$\Longleftrightarrow \ln|z| + i\mathrm{Arg}(z) = i$$
\begin{equation}
\Longleftrightarrow 
\begin{cases}
\ln|z| = 0\\
\arctan(y/x) = 1
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow 
\begin{cases}
|z| = 1\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow 
\begin{cases}
x^2 + y^2 = 1\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow 
\begin{cases}
x^2 = 1/2\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow 
z = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}
\end{equation}
Where is the problem in my calculations ?


Assuming $x + y \cdot \mathrm{i} := z$, the right answer is $\cos(1) + \sin(1) \cdot \mathrm{i} $. I found: [...]
Where is the problem in my calculations?

The problems are the step from $\arctan\left( \frac{y}{x} \right) = 1$ to the next one and $x = y$.
The argument is not always the arctangent of $\frac{y}{x}$, e.g. when $x = 0$! In addition, $x \ne y$! Accordingly, their premises are wrong and so is the calculation.
E.G. If you say $\arctan\left( \frac{y}{x} \right) = 1$ and $x = y$ it follows:
$$
\begin{align*}
\arctan\left( \frac{y}{x} \right) &= 1\\
\arctan\left( \frac{y}{y} \right) &= 1\\
\arctan\left( 1 \right) &= 1\\
\frac{\pi}{4} + 2 \cdot k \cdot \pi &= 1\\
\end{align*}
$$
And $1$ is not the same as $\frac{\pi}{4} + 2 \cdot k \cdot \pi$!
Correct step by step solution with your method
If you go through it correctly you will get:
$$
\begin{align*}
\ln(z) &= \mathrm{i} \quad\mid\quad \text{use polar form } z = |z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i})\\
\ln(|z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i})) &= \mathrm{i} \quad\mid\quad \text{find } |z| \text{ and find } \arg{z}\\
\ln(|z|) + \arg(z) \cdot \mathrm{i} &= \mathrm{i}\\
\begin{cases} (1.) \qquad \ln(|z|) &= 0 \\ (2.) \qquad \arg(z) \cdot \mathrm{i} &= \mathrm{i}\end{cases}\\
(1.) \qquad \ln(|z|) &= 0 \quad\mid\quad \exp(~~)\\
(1.) \qquad |z| &= 1\\
(2.) \qquad \arg(z) \cdot \mathrm{i} &= \mathrm{i} \quad\mid\quad \div \mathrm{i}\\
(2.) \qquad \arg(z) &= 1\\
\\
z &= |z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i}) \quad\mid\quad |z| = 1 \wedge \arg(z) = 1\\
z &= 1 \cdot (\cos(1) + \sin(1) \cdot \mathrm{i})\\
z &= \cos(1) + \sin(1) \cdot \mathrm{i}\\
\end{align*}
$$
But there is also a much easier way to solve for $z$:
$$
\begin{align*}
\ln(z) &= \mathrm{i} \quad\mid\quad \exp(~~)\\
\exp(\ln(z)) &= \exp(\mathrm{i})\\
z &= \exp(\mathrm{i})\\
z &= \operatorname{cis}(1) \quad\mid\quad \text{use euler's formula } \operatorname{cis}(x) =  \cos(x) + \sin(x) \cdot \mathrm{i}\\
z &= \cos(1) + \sin(1) \cdot \mathrm{i}\\
\end{align*}
$$
