Proving $\bar z_1+\bar z_2=\overline{z_1+z_2}$ and other $\overline{z}_1+\overline{z}_2=\overline{z_1+z_2} $
Can I just assign two random values and prove it that way? (My textbook doesn't have an answer.) 
I also have these problems, but I don't understand them either :( 
$\frac{\overline{z}_1}{\overline{z}_2}=\overline{\left(\frac{z_1}{z_2}\right)}$ 
$\overline{z}_1\:\overline{z}_2=\overline{z_1z_2}$ 
If anyone can help, I would really appreciate it!
 A: Hint for Method 1 (as told by Nameless) (algebraic)
Let $z = x + iy$ 
hence,
$\overline{z} = x - iy$ 
Now, try to work out each case.

Hint for Method 2 (geometrical)
Use the following facts, 


*

*Complex numbers can be represented on the Argand plane.

*Taking the modulus of a complex number is interpreted geometrically as taking its refection on the $X$ axis (real axis).

*Adding two complex numbers is just like adding two vectors.

*Multiplication of two complex numbers geometrically involves multiplying their magnitudes and adding their arguments.



If you are still stuck or if you are only able to prove by one method, please comment and I will provide a detailed solution with both methods.
A: We can't just pick and random values as this statement is about every complex number, not just two of them.
So instead, lets right $z_1$ as $x_1+iy_1$ and $z_2$ as $x_2+iy_2$
Note that if $z=x+iy$ is a complex number, the definition of $\bar{z}$ is $x-iy$
So the left hand side can be written as $\bar{z_1}$+$\bar{z_2}$=$\overline{x_1+iy_1}+\overline{x_2+iy_2}$=$x_1-iy_1$+$x_2-iy_2$
Collecting like terms, we get that the LHS is equal to $x_1+x_2-iy_1-iy_2$
Now can you figure out how to simplify the RHS as well from this?
How can you show they are equal?
A: The problem you are talking is about the Well-Definition of an equality in maths.
At this link you can see what is the definition of "well defined" in mathematics.
First of all we have $\exists a,b,c,d\in\mathbb{R}$ such as $$\mathbb{C}\ni z_1=a+ib\implies\overline{z}_1=a-ib$$ $$\mathbb{C}\ni z_2=c+id \implies\overline{z}_2=c-id $$
Now let’s show that your formulas are well defined:

$$\overline{z}_1+\overline{z}_2=\overline{z_1+z_2}\tag{1} $$
Then calculate the sum $$z_1+z_2=a+ib+c+id=(a+c)+i(b+d)\implies \\\overline{z_1+z_2}=\overline{(a+c)+i(b+d)}=(a+c)-i(b+d)=(a-ib)+(c-id)=\bar{z_1}+\bar{z_2}$$

$$\overline{z}_1\overline{z}_2=\overline{z_1z_2}\tag{2} $$
Then calculate the product $$z_1z_2=(a+ib)(c+id)=ac+ida+ibc+i^2bd\overbrace{=}^{i^2=-1}(ac+bd)+i(ad+bc)\implies \\\overline{z_1z_2}=\overline{(ac+bd)+i(ad+bc)}=(ac+bd)-i(ad+bc)=(a-ib)(c-id)=\bar{z_1}\bar{z_2}$$

$$\frac{\overline{z}_1}{\overline{z}_2}=\overline{\left(\frac{z_1}{z_2}\right)}\tag{3}$$
Then calculate the division:
$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{\frac{ac+bd}{c^2+d^2}+i \frac{cb -ad}{c^2+d^2}}=\frac{ac+bd}{c^2+d^2}-i \frac{cb -ad}{c^2+d^2}=\\=\frac{ac+(-b)(-d)}{c^2+(-d)^2}+i\frac{c(-b)-a(-d)}{c^2+(-d)^2}=\frac{a-ib}{c-id}=\frac{\bar{z_1}}{\bar{z_2}}$$
$$\tag*{$\square$}$$
