Can all complex expressions be simplified to the form $a+jb$? Are there any complex expressions that cannot be simplified to the form $a+jb$, where a and b are real numbers?
For example, $$\frac{1}{j}=0+j(-1),\hspace{0.5cm}e^j=\cos(1)+j\sin(1),\hspace{0.5cm}\sin(j)=0+j\frac{e^2-1}{2e}$$
From what I understand, all complex numbers must exist somewhere on the complex plane where a and b are the coordinates. But some expressions don't have any obvious way to be simplified:
$$\ln(1+j)=???,\hspace{0.5cm}\arctan(j)=???$$
If every expression can be simplified, are there any good references or list of identities?
 A: To get
$\ln(z)$
for any complex $z$
(where $z \ne 0$),
write
$z = |z|y$.
Then $|y| = 1$,
so there is a real $t$
such that
$y = e^{it}
=\cos(t)+i\sin(t)
$,
so
$z 
= |z|y
= |z|(\cos(t)+i\sin(t))
= |z|\cos(t)+i|z|\sin(t)
$.
Since
$|1+j| = \sqrt{2}$,
$(1+j)
=\sqrt{2}\frac{1+j}{2}
=\sqrt{2}e^{i\pi/4}
$,
so
$\ln(1+j)
=\ln(\sqrt{2}e^{i\pi/4})
=\dfrac{\ln(2)}{2}+i\pi/4
$
(you can throw in
$+2\pi  i n$
if you want).
A: Yes, all complex expressions that produce a complex number ($\arctan(i)$ doesn't) can be written in rectangular form. You can generally use a Taylor Series or convert your expression to another form to find the complex number in rectangular form. Here is a page that deals with a lot of these cases. Most of these results use $e^{ix}=\cos(x)+i\sin(x)$ to convert functions that don't seem to work with complex numbers into ones that do.
A: Assume $z=re^{i\theta}=x+iy$. Then
$$\ln(z)=\ln(r)+\ln(e^{i\theta})=\ln(r)+i\theta=\ln(\sqrt{x^2+y^2})+i\arctan(\frac{y}{x})$$
Also regarding $\arctan(z)$, it is not difficult to show that $$\arctan(z)=\frac{1}{2i}\ln\left(\frac{i-z}{i+z}\right)$$
A: The first one is simple:
$$\ln(1+j) = \ln(\sqrt{2} e^{j\pi/8}) = \frac12 \ln 2 + j\frac\pi8$$
For the second: if
$$z = \tan w = \frac{\sin w}{\cos w} = \frac1j \frac{e^{jw} - e^{-jw}}{e^{jw}+e^{-jw}}$$
we can solve for $w = \arctan z$:
$$ e^{2jw} = \frac{1+jz}{1-jz} $$
$$\arctan(z) = w = \frac12 \ln e^{2jw} = \frac12 \ln \left( \frac{1+jz}{1-jz} \right).$$
In particular when $z = j$:
$$\arctan(j) = \frac12 \ln \left( \frac{1+j^2}{1-j^2} \right) = \frac12 \ln 0 $$
and, well, $\ln 0$ is undefined.
A: Any complex number $z$ can be expressed as $Re(z)+i Im(z)$.
