$\arg(z_1+z_2)=0$ and $\text{Im}(z_1z_2)=0$. Prove that $z_1=\bar{z_2}$ 
$\arg(z_1+z_2)=0$ and $\text{Im}(z_1z_2)=0$. Prove that $z_1=\bar{z_2}$

I substituted $z_1=a+ib$ and $z_2=c+id$
Using first condition,
$$z_1+z_2=\bar{z_1}+\bar{z_2}$$
$$a+ib+c+id=a-ib+c-id$$
$$b+d=0$$
So $z_1=a+ib$, $z_2=c-ib$
Using second condition, 
$$\text{Im}(ac-iab+ibc+b^2)=0$$
If $b\neq 0$
$$a=c$$
So $z_1=a+ib$ and $z_2=a-ib$
Is there a way without substitution?
 A: Saying that $\arg(z_1+z_2)=0$ means that $z_1+z_2$ is real and nonnegative. In particular the imaginary parts sum to $0$ and for the real parts you have $a+c\ge0$.
The second condition says $ad+bc=0$, which together with $d=-b$ tells you $b(a-c)=0$. Thus either $b=0$ (so also $d=0$), or $a=c$.
Thus, the statement is false, because obviously $z_1=1$ and $z_2=2$ satisfy the requirements, but are not conjugate of each other.
Therefore, no: there no way to prove the result, with or without substitutions.
If one assumes $z_1$ and $z_2$ nonreal, the result is true. The first condition implies $z_1+z_2=\bar{z}_1+\bar{z}_2$, the second condition that $z_1z_2=\bar{z}_1\bar{z}_2$. Therefore
$$
0=z_1z_2-\bar{z}_1\bar{z}_2=
z_1z_2-z_1\bar{z_2}+z_1\bar{z_2}-\bar{z}_1\bar{z}_2=
z_1(z_2-\bar{z}_2)+(z_1-\bar{z}_1)\bar{z}_2
$$
Applying the fact that $z_2-\bar{z}_2=-(z_1-\bar{z}_1)$, we get
$$
(z_1-\bar{z}_1)(z_1-\bar{z}_2)=0
$$
Since $z_1-\bar{z}_1\ne0$, we deduce $z_1-\bar{z}_2=0$.
A: Since $arg(z_1+z_2)=0$ you have
$$z_1+z_2=\bar{z_1}+\bar{z_2}=:a$$
Since $Im(z_1z_2)=0$ you have
$$z_1z_2=\bar{z_1} \bar{z_2}=:b$$
Now, consider the quadratic equation 
$$z^2-az+b=0$$
Then $z_1,z_2$ are the two roots. Also $\bar{z_1}, \bar{z_2}$ are the two roots.
This proves that 
$$z_1=\bar{z_1} \,;\, z_2 =\bar{z_2} \mbox{ or } \\
z_1=\bar{z_2} \,;\, z_2 =\bar{z_1} $$
The first case provides a counterexample, but under the extra assumption that $z_1$ is not real, you get the desired conclusion.
