# Number of z, which satisfy Arg(z – 3 – 2i) = $\frac{\pi}{6}$ and Arg(z – 3 – 4i) = $\frac{\pi}{3}$ are?

I plugged in the values in Wolfram Alpha and found that there aren't any solution for z.

But i tried by putting $z=x+yi$ in both equations and i'm getting one solution.

For the first condition, i used $$\frac{y-2}{x-3} = \tan{\frac{\pi}{6}}$$ from which i get one equation $$y=\frac{x}{\sqrt{3}}-\sqrt{3}+2$$

Similarly for the second condition, i get the second equation as, $$y=\sqrt{3}x-3 \sqrt{3} + 4$$

On solving the two equations, i'm getting one solution of x and y, which gives one solution for z. Is there anything i'm doing wrong ? I'm kind of weak in complex numbers

• these are part-lines not complete lines, which is why WA gives no solution – David Quinn Mar 22 '18 at 12:18
• You’ve come up with a linear system of two variables, so it has a unique solution. – Riccardo Ceccon Mar 22 '18 at 12:29
• oh, ok. I tried solving this using pure geometry only, so how do i solve it ? – Harshit Mar 22 '18 at 12:55
• Did you get $z=\left(3-\sqrt{3}\right)+i$ ? Replace in the original equations to see if $\arg (z-3-2i)= \pi /6$ – Lozenges Mar 22 '18 at 12:55
• Oh, so i stopped in between getting the solution, that means my method is correct ? @Lozenges yes i got that only. – Harshit Mar 22 '18 at 13:07

The easiest way to see that there is no solution is to sketch these part-lines in an Argand Diagram. The first is a line going from $3+2i$ at angle $\frac{\pi}{6}$ to the positive real axis, and similarly for the second line.