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.

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. 
