Vrftsjtryk

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

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 $tantheta = frac{text{imaginary part}}{text{real part}}$

$z-3-4i = (x-3)+(y-2)i$

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 $x= 3+frac{5sqrt3}{2}-2sqrt3$

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 $$arg(z-3-4i)=frac{2pi}{3}$$ 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.