Proof of equality of complex numbers with conditions 
I have the next question:
 Proof that $\operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=|z_{1}|\cdot |z_{2}|$ only if:  $\arg z_{1} -\arg z_{2}  =2\cdot \pi \cdot n$ ( $\;(n\in \mathbb{Z})$. 
 This is my solution:
$z_{1}=x+iy,   z_{2}=a+ib$
$z_{1}\cdot \bar{z_{2}}=(x+iy)\cdot (a-ib)=xa+yb+i(ya-bx)$
$\rightarrow \operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=xa+yb$
 $|z_{1}|=\sqrt{x^{2}+y^{2}}, |z_{2}|=\sqrt{a^{2}+b^{2}}$$|z_{1}|\cdot |z_{2}|=\sqrt{(x^{2}+y^{2})\cdot (a^{2}+b^{2})}=\sqrt{x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}}$

$\Rightarrow (xa+yb)=\sqrt{x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}}$
$\overset{()^{2}}{\rightarrow}x^{2}a^{2}+2xayb+y^{2}b^{2}=x^{2}a^{2}+x^{2}b^{2}+y^{2}a^{2}+y^{2}b^{2}$
$\rightarrow2xyab=x^{2}b^{2}+y^{2}a^{2}\rightarrow x^{2}b^{2}-2xyab+y^{2}a^{2}=0\rightarrow (xb-ya)^2=0$

$x=r_{1}\cos\theta , y=r_{1}\sin\theta$
$a=r_{2}\cos\varphi  , b=r_{2}\sin\varphi $
$(r_{1}\cos\theta \cdot r_{2}\sin\varphi-r_{1}\sin\theta\cdot r_{2}\cos\varphi)^2=0$
$(r_{1} \cdot r_{2} (\cos\theta \sin\varphi-\sin\theta \cos\varphi))^2=0$
$\sin^2(\varphi-\theta)=0$


For every $n$ that we put in the condition, the equation is true but the angle $\pi$ is also a solution for this equation but it is not created from the condition.
 Where have I done a mistake? 
 Thank you all for the help.
 A: A better way to solve the problem (not addressing where your solution might be wrong): write $z = re^{i\theta}$ and $w = se^{i\phi}$ and calculate $z \bar{w}$.
This is easy if you know about complex exponentials. If you don't, think of
$$
e^{i\theta} = \cos \theta + i \sin \theta
$$
as an abbreviation.
A: Let $\mathbf{z}$ denote the vector $\in\mathbb{R}^2$ corresponding to $z$ so if $z=x+i y$, $\mathbf{z}=\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$. 
Then $\operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=\mathbf{z_1}.\mathbf{z_2}$ i.e. the dot product in $\mathbb{R}^2$. This is because $$\operatorname{Re}\{(x_1+iy_1).(x_2-iy_2)\}=\operatorname{Re}\{x_1x_2+y_1y_2+i(y_1x_2-x_1y_2)\}=\left(\begin{smallmatrix}x_1\\y_1\end{smallmatrix}\right)\cdot \left(\begin{smallmatrix}x_2\\y_2\end{smallmatrix}\right)$$
Hence the original equation $$\operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=|z_{1}|\cdot |z_{2}|$$  is equivalent to 
$$\mathbf{z_1}.\mathbf{z_2}=|z_1|\cdot |z_2|=|\mathbf{z_1}|\cdot |\mathbf{z_2}|$$ But $$\mathbf{z_1}.\mathbf{z_2}=|\mathbf{z_1}|\cdot |\mathbf{z_2}| \cos{\theta}$$ where $\theta$ is the angle between the vectors $\mathbf{z_1}$ and $\mathbf{z_2}$. Hence $\cos{\theta}=1$, meaning that the vectors point in the same direction and $\mathbf{z_1}=\lambda\mathbf{z_2}$ with $\lambda \in \mathbb{R}$ and $\lambda> 0$. This implies, translating back to complex numbers, that the arguments of the complex numbers differ by a full rotation or $2\pi$ radians:
  $$\arg z_{1} -\arg z_{2}  =2\cdot \pi \cdot n$$
A: Hint: 
Condition $\arg z_{1} -\arg z_{2}  =2\cdot \pi \cdot n$ $\;(n\in \mathbb{Z})$ is the same as writing $z_2 = k\cdot z_1$ for some real positive number $k$.
Also recall $$Re(z) = {z+\bar{z}\over 2}\;\;\;\;\;\;{\rm and}\;\;\;\;\;\;|z|^2 = z\cdot \bar{z}$$ 
Since $$Re(z_1\bar{z}_2) = {z_1\bar{z}_2+z_2\bar{z}_1\over 2}\;\;\;\;\;\;{\rm and}\;\;\;\;\;\;|z_1z_2|^2 = z_1z_2\cdot \bar{z_1} \bar{z_2}$$ 
we have $$(z_1\bar{z}_2+z_2\bar{z}_1)^2 = 4z_1z_2\cdot \bar{z_1} \bar{z_2}$$ which implies $$(z_1\bar{z}_2-z_2\bar{z}_1)^2 =0\implies z_1\bar{z_2} = z_2\bar{z_1}\implies $$ $$z_1\cdot \bar{z_2}= |z_1z_2|\implies z_1 =z_2\cdot \underbrace{\Big|{z_1\over z_2}\Big|}_k$$
$$z_1 = k\cdot z_2\implies  \arg(z_1)-\arg(z_2)=2\pi n$$
A: There is a hidden thing you should be careful of when you square both sides you assuming  that $ax+by>0$ so the equality make sense . But this thing means $\cos \phi \cos \theta +\sin \phi  \sin \theta >0$ hence $\cos(\theta -\phi)>0$ so the difference is either in first or fourth quarter but if the difference is multiple of $\pi$ then it could happen one solution in first then the other would be in the third. 
