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Solve geometrically the following equation : $arg(z) +arg(z+1)=0 \pmod{\pi}$

My solution: Let: $z=x+iy$
Then: $\arg(z)+\arg(z+1)=0 \pmod{\pi} \Longleftrightarrow z(z+1) \in \mathbb{R}$
Then: $\operatorname{Im}((x+iy) (x+1+iy))=0$
After simplification we get $x=-1/2 $ or $ y=0$

However, that's an algebraic method and I'm required to solve it geometrically. Any ideas?

Neptune
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Oussama Akar
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1 Answers1

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Geometrically speaking, if $z$ is in the first or fourth quadrant, then $z(z+1)$ can never lie on the $x-$axis (i.e., $z(z+1)$ can never be real). E.g. if $z$ is in the first quadrant, then $\arg(z)\in(0,\pi/2)$ and $\arg(z+1)\in(0,\pi/2)$ with $\arg(z+1)<\arg(z)$, so $\arg(z)+\arg(z+1)\in(0,\pi)$.

Notice that we can show the desired $z$ does not lie on $y-$axis (except for the origin) either with a similar way. And if $z$ lies on the $x-$axis, then $z(z+1)$ is obviously real.

Now, suppose $z$ is in the second or the third quadrant. Say $z$ lies in the second quadrant (the third quadrant case is very similar).

  • Case 1: Suppose $z+1$ lies also in the second quadrant. Then $\arg(z)+\arg(z+1)\in(\pi,2\pi)$, so $z(z+1)$ is not real.
  • Case 2: Suppose $z+1$ lies in the first quadrant. Then $\arg(z)+\arg(z+1)\in(\pi/2,3\pi/2)$. Thus, if $z(z+1)$ is real, then we must have $\arg(z)+\arg(z+1)=\pi$. This means $z$ and $z+1$ must be symmetric w.r.t. $y-$axis. Hence $Re(z)=-1/2$.
Yuqiao Huang
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