Solve for $z$, which satisfy $\arg(z-3-2i) = \frac{\pi}{6}$ and $\arg(z-3-4i) = \frac{2\pi}{3}$.

Question

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?

Answer 1

$$\arg(z-3-2i)=\frac{\pi}{6}$$ 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.