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

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

Clearly they do not meet.

Alternatively establish the point of intersection of the two Cartesian lines and check that in the first case the real part must be greater than 3 and the imaginary part greater than 2. Likewise for the second case. Again no solution.


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