Polar form of $z^2+1$, $z\in \mathbb C$ Let $z\in \mathbb C$. Can we write $z^2+1 = re^{i\theta}$, for some $r\in \mathbb R$ and $\theta \in (-\pi, \pi]$?
We have $z^2+1 = r^2e^{i2\theta}+e^{i0}$ or we can write $z^2+1 = x^2-y^2+1 + i2xy$. We can use formulas to calculate $r$ and $\theta$ of $z^2+1$, but the process seems to be long. From the geometric point of view, I think $z^2+1$ is stretching the length of $z$ to $r^2$ and rotating the angle $\theta$ to $2\theta$, then shifting the vector to the right by a unit.
My ultimate goal is to calculate $\frac{d}{dz}Log(z^2+1)$. I think we can write $Log(z^2+1)$ in terms of $x$ and $y$, then use the formula $f'(z) = u_x + iv_x$, but I would like to see if I can use $f'(z) = e^{-i\theta}(u_r + iv_r)$. 

First of all, notice that $z^{2} + 1 = (z-i)(z+i)$. Thus we have:
\begin{align*}
|z^{2} + 1| = |z-i||z+i|
\end{align*}
If we agree that $z = \rho e^{i\theta}$, we get
\begin{align*}
z - i = \rho\cos(\theta) + i(\rho\sin(\theta) - 1) \Rightarrow |z - i|^{2} & = \rho^{2}\cos^{2}(\theta) + (\rho\sin(\theta) - 1)^{2}\\
& = \rho^{2} - 2\rho\sin(\theta) + 1
\end{align*}
Similarly, we have:
\begin{align*}
z + i = \rho\cos(\theta) + i(\rho\sin(\theta) + 1) \Rightarrow |z+i|^{2} & = \rho^{2}\cos^{2}(\theta) + (\rho\sin(\theta) + 1)^{2}\\
& = \rho^{2} + 2\rho\sin(\theta) + 1
\end{align*}
From whence we obtain:
\begin{align*}
|z^{2} + 1|^{2} = |z-i|^{2}|z+i|^{2} = (\rho^{2} + 1)^{2} - 4\rho^{2}\sin^{2}(\theta)
\end{align*}
Based on this, it is possible to find the argument of $z^{2} + 1$ as well, since we have
\begin{align*}
\begin{cases}
\displaystyle\cos(\theta_{1}) = \frac{\mathrm{Re}(z-i)}{|z - i|} = \frac{\rho\cos(\theta)}{\sqrt{\rho^{2} - 2\rho\sin(\theta)+1}}\\
\displaystyle\cos(\theta_{2}) = \frac{\mathrm{Re}(z+i)}{|z+i|} = \frac{\rho\cos(\theta)}{\sqrt{\rho^{2}+2\rho\sin(\theta)+1}}
\end{cases}
\end{align*}
\begin{align*}
\therefore z^{2} + 1 = \sqrt{(\rho^{2}+1)^{2} - 4\rho^{2}\sin^{2}(\theta)}\times e^{i\alpha}\,\,\text{where}\,\,\alpha = \theta_{1} + \theta_{2}
\end{align*}
