Solve for $z$, which satisfy $\arg(z-3-2i) = \frac{\pi}{6}$ and $\arg(z-3-4i) = \frac{2\pi}{3}$. solve for $z$, which satisfy $\displaystyle \arg(z-3-2i) = \frac{\pi
}{6}$ and $\displaystyle\arg(z-3-4i) = \frac{2\pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting  real parts together and imaginary parts together.
Then I'm using $\tan\theta = \frac{\text{imaginary part}}{\text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$\tan30° = \frac{y-2}{x-3}$
$\frac{1}{\sqrt3} = \frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+\frac{5\sqrt3}{2}-2\sqrt3$
But the answer key says there's no such $z$ which satisfies this equation. 
Is this the wrong way to solve this question, is my answer wrong or the answer key's? 

$$\arg(z-3-2i)=\frac{\pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^\circ$ with positive $x$ axis.
And $$\arg(z-3-4i)=\frac{2\pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^\circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis. 
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
