Prove that the following are real numbers 
  
*
  
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
  
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
  

1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z)}{|z|}$$ which is real number as a division of two real numbers


*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$


$$\overline{z^3\cdot\overline{z}+z\cdot\overline{z}^3}$$
Due to: $\overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$ and $\overline{z_{1}\cdot z_{2}}=\overline{z_{1}}\cdot\overline{z_{2}}$
we get:
$$\overline{z}^3\cdot{z}+\overline{z}\cdot{z}^3$$
Now we can say that we got the same elements and $(z=\overline{z})$ therefore it is real number? 
Is the reasoning in both of these correct?
 A: Alternatively:


*

*$\cfrac{1}{z}+\cfrac{1}{\overline{z}}=\left(\cfrac{1}{z}\right)+\overline{\left(\cfrac{1}{z}\right)}= 2 \operatorname{Re}\left(\cfrac{1}{z}\right)$

*$z^3\cdot\overline{z}+z\cdot\overline{z}^3 = z \bar z(z^2 + \bar z ^2) = |z|^2\cdot 2 \operatorname{Re}(z^2)$
A: Put $z=re^{it} $ with $r\neq 0$.
then
$$\frac 1z+\frac {1}{\overline{z}}=$$
$$\frac {1}{r}(e^{it}+e^{-it})=\frac {2\cos (t)}{r }\in\mathbb R $$
and
$$z^3\overline {z}+z\overline {z}^3=$$
$$r^4 (e^{2it}+e^{-2it})=r^4\cos (2t) \in\mathbb R $$
A: Intuitively/geometrically, inverting a complex number means reversing its angle and stretching/shrinking its radius:
$$\frac1{z} = \frac1{re^{i\theta}} = \frac1r e^{-i\theta}$$
Since conjugation is just reversing the angle ($\bar{z} = e^{-i\theta}$), when we add the inverse of $z$ and its conjugate, the angle will be 0 (since the angle of $\frac1{z}$ will be $-\theta$ and the angle of $\frac1{\bar{z}}$ will be $\theta$), so it's real. 

For the second expression, first factor it as:
$$z^3 \bar{z} + z \bar{z}^3 = z\bar{z}\left( z^2 + \bar{z}^2 \right)$$
The first element of the product is obviously real.
Squaring a complex number means doubling its angle. So when we add the square of $z$ to the square of its conjugate, the angles will cancel out. 
So the second element of the product is real as well. And the product of reals is real.
A: If we know $z + \overline z \in \mathbb R$ and $z*\overline z = |z|^2\in \mathbb R$ and $\overline{\overline z} = z$ and $\overline {z*w} = \overline z*\overline w$   we are pretty much done.
1) $\frac 1{\overline z} *\overline z = 1$
$\overline{\frac 1{\overline z}*\overline z} = \overline 1 = 1$
$\overline{\frac 1{\overline z}}*\overline{\overline z} = \overline{\frac 1{\overline z}}*z = 1$
$\overline{\frac 1{\overline z}} = \frac 1z$ so 
$\frac {1}{\overline z} = \overline {\frac 1z}$ so
$\frac 1z + \frac 1{\overline z} = \frac 1z + \overline {\frac 1z} \in \mathbb R$.
2) $z^3*\overline z +z*\overline{z^3}= \overline{\overline {z^3}}*\overline z +\overline{z^3}*z= \overline{\overline {z^3}*z}+\overline{z^3}*z\in \mathbb R$
A: we have $$\frac{z+\overline{z}}{z\overline{z}}=\frac{2x}{x^2+y^2}$$ if $$z=x+iy$$
