Why are complex numbers allowed to be combine like this? The problem
This expression is meant to be simplified. Why does it make sense to write $$\frac{1}{1-6i} - \frac{1}{1+6i} = \frac{1+6i−(1−6i)}{(1−6i)(1+6i)} \quad ?$$
How can the rules followed here be applied to similar problems?
 A: Here's how you simplify
$$
\frac{1}{1-6i}-\frac{1}{1+6i}.
$$
You multiply the numerator and denominator of each fraction by the conjugate of the complex number (change the sign of the part with $i$).  In other words, you have
$$
\frac{1+6i}{(1+6i)(1-6i)}-\frac{1-6i}{(1+6i)(1-6i)}=\frac{1+6i}{1+36}-\frac{1-6i}{1+36}=\frac{12i}{37}.
$$
This is your right-hand-side. (the original post had a very different RHS).
A: 
$$\frac{1}{1-6i}-\frac{1}{1+6i}$$

$$=\frac{1}{1-6i}\cdot\color{blue}{\frac{(1+6i)}{(1+6i)}}-\frac{1}{1+6i}\cdot\color{blue}{\frac{(1-6i)}{(1-6i)}}$$
$$=\frac{1+6i}{37}-\frac{1-6i}{37}=\frac{1+6i-1+6i}{37}=\boxed{\frac{12}{37}i}$$
A: Notice, when $z\in\mathbb{C}$:


*

*$$\overline{z}=\frac{|z|^2}{z}\Longleftrightarrow\frac{1}{\overline{z}}=\frac{z}{|z|^2}$$


Now, in your example, set $z=1+6i$:
$$\frac{1}{\overline{z}}-\frac{1}{z}=\frac{z}{z\overline{z}}-\frac{\overline{z}}{z\overline{z}}=\frac{z-\overline{z}}{z\overline{z}}=\frac{z-\frac{|z|^2}{z}}{z\cdot\frac{|z|^2}{z}}=\frac{z-\frac{|z|^2}{z}}{|z|^2}=\frac{z}{|z|^2}-\frac{1}{z}$$
And so:
$$\frac{1+6i}{|1+6i|^2}-\frac{1}{1+6i}=\frac{1+6i}{37}-\frac{1-6i}{37}=\frac{1+6i-1+6i}{37}=\frac{12i}{37}$$
A: $$
\frac{1}{\overline{z}}-\frac{1}{z}
=\frac{z-\overline{z}}{z\overline{z}}
=\frac{2Im(z)}{|z|^2}
=\dfrac{12i}{37}
\qquad\text{for} \ z=1+6i
$$
