# Does $\pi/4=\arctan\left(\frac{-y}{x^2+y^2-5x+6}\right)$ imply $x^2+y^2-5x\pm y+6=0$

While solving $$\arg\left(\frac{z-2}{z-3}\right)=\pi/4$$, I arrive at the following equation $$\pi/4=\arctan\left(\frac{-y}{x^2+y^2-5x+6}\right)$$

At this point I take $$\tan()$$ on both sides to get $$x^2+y^2-5x+y+6=0$$. But according to my book the answer is $$x^2+y^2-5x\pm y+6=0$$. I possibly made a mistake in the simplification of the $$\arctan()$$ equation. What is it?

Although I'm not sure what your reasoning is, and I'm now starting to doubt my own reasoning, I agree with your conclusion, and I disagree with the book, whatever it is! Perhaps we're both making the same mistake? Let's see.

The fundamental problem is that for $$u + iv \in \mathbb{C},$$ with $$u \ne 0,$$ the equation: $$$$\label{3460069:eq:1}\tag{1} \arg(u + iv) = \frac\pi4$$$$ is not equivalent to: $$$$\label{3460069:eq:2}\tag{2} \arctan\frac{v}{u} = \frac\pi4.$$$$ Equation \eqref{3460069:eq:2} is - of course! - equivalent to: $$$$\label{3460069:eq:2p}\tag{2'} \frac{v}{u} = 1, \text{ i.e. } u = v.$$$$ If $$u = v > 0$$, then \eqref{3460069:eq:1} does hold, but on the other hand, if $$u = v < 0$$, we have: $$$$\label{3460069:eq:3}\tag{3} \arg(u + iv) = -\frac{3\pi}4.$$$$ In sum: either of \eqref{3460069:eq:2} or \eqref{3460069:eq:2p} is equivalent to the exclusive disjunction of \eqref{3460069:eq:1} and \eqref{3460069:eq:3}.

In the present problem, you presumably arrived at some condition such as: $$\arg((z - 2)(\overline{z} - 3)) = \frac\pi4$$ which, for $$z = x + iy$$, is equivalent to: $$\arg(x^2 + y^2 - 5x + 6 - iy) = \frac\pi4.$$ By my apparently simple reasoning, from the above general considerations, this implies $$y < 0$$, and also implies: $$x^2 + y^2 - 5x + y + 6 = 0.$$

Are we both wrong?

Geometrical common sense (call it "intuition", to be posh) does seem to confirm that $$z - 2$$ can only be an anticlockwise quarter-turn away from $$z - 3$$ if $$z$$ is in the lower half-plane $$\{x + iy : y < 0\}$$, and this is somewhat reassuring.

If I've screwed up, I'll post a correction, but for the moment I'm inclined to suspect there's an error in the book.

(I should add that the problem in the title of the question differs from the problem stated in the question itself. Rather than risk adding to the confusion, I ask you to confirm what exactly is the question in the book, and modify the title and/or body of the question appropriately.)

## Update

The figure illustrates the proposition that the locus of points $$z$$ in the complex plane such that: $$\arg\left(\frac{z - 2}{z - 3}\right) = \frac\pi4$$ is the intersection of the circle $$\{ x + iy : x^2 + y^2 - 5x + y + 6 = 0 \}$$ with the lower half-plane $$\{x + iy : y < 0\}.$$

The small arc of the same circle in the upper half-plane $$\{x + iy : y > 0\}$$ is associated with the angular value $$-\frac{3\pi}4.$$

For completeness I've also shown arcs of the circle $$\{ x + iy : x^2 + y^2 - 5x - y + 6 = 0 \}$$ and their associated angular values $$-\frac\pi4$$ and $$\frac{3\pi}4.$$

If you've quoted the book accurately, there's definitely an error in it.

• Wait why is $(1)$ not equivalent to $(2)$? Is it because of the possibility you mention of this being $-3\pi/4$. – Paras Khosla Dec 2 at 19:32
• Alright. Thanks – Paras Khosla Dec 2 at 19:32