How to break $\ln(x+i \cdot y)$ into real and imaginary components? Wolfram shows that there are indeed Re and Im components without stating them: http://www.wolframalpha.com/input/?i=ln(x%2Bi*y)
How do I break $\ln(x+i \cdot y)$ into real and imaginary components?
 A: We want to express $x+iy$ as $re^{i\theta}$, for some value of $r$ and $\theta$. 
First let's normalize $x+iy$ as $\sqrt{x^2+y^2}(\frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}i)$.
We see that $r=\sqrt{x^2+y^2}$ and we see that $\tan(\theta)=\frac{y}{x}$. Therefore, we have $\theta=\arctan(\frac{y}{x})$. We can then write $x+iy=\sqrt{x^2+y^2}e^{i\arctan(\frac{y}{x})}$. So we have:
$$\ln(x+iy)=\ln(\sqrt{x^2+y^2}e^{i\arctan(\frac{y}{x})})$$
$$\ln(x+iy)=\frac{1}{2}\ln(x^2+y^2)+i\arctan(\frac{y}{x})$$
Our real part is $\frac{1}{2}\ln(x^2+y^2)$ and our imaginary part is $i\arctan(\frac{y}{x})$.
A: There are different ways to introduce a complex logarithm, but the definition which is used by WolframAlpha is
$$\ln(z) := \ln |z| + i \text{Arg} z$$
(see (8) here) where
$$\text{Arg} \, z = \arctan \left( \frac{y}{x} \right)$$
denotes the argument of $z=x+iy \in \mathbb{C}$ (see (2) here). Hence,
$$\ln(x+iy) = \ln \sqrt{x^2+y^2} + i \arctan \left( \frac{y}{x} \right)$$
which implies
$$\text{Re} \, \ln(x+iy) = \ln \sqrt{x^2+y^2}$$
and
$$\text{Im} \, \ln(x+iy) = \arctan \left( \frac{y}{x} \right).$$
