Finding $z \in \mathbb{C}$, with $z \neq -1, i$, such that $\operatorname{arg}(z+1)=\operatorname{arg}(z-i)$.

Question

So for most locus questions, the algebra is fairly straightforward, I usually just substitute $z = x + yi$ into $z$ and see where that takes me. But for this question, I can't seem to get anywhere with that method and I can't seem to find a intuitive way of visualizing it either so how would I go about finding the locus of this equation?

Problem:

Find the set of all $z \in \mathbb{C}$ with $z \neq -1,i$ such that $\operatorname{arg}(z+1)=\operatorname{arg}(z-i)$.

Any help would be greatly appreciated!

Answer 1

Since the argument of a complex number is real whenever it's defined, we can write the locus as $$\bigcup_{r\in \mathbb{R}}(\{z : \operatorname{arg}(z+1) = r\} \cap \{z : \operatorname{arg}(z-i)=r\})$$ (in words, the argument of $z+1$ equals the argument of $z-i$ if and only if they both equal the same real number)

The benefit is that the loci involved here are geometrically nice. Specifically, $$\{z : \operatorname{arg}(z+1) = r\} \\ \{z : \operatorname{arg}(z-i)=r\}$$ are two rays with endpoints at $-1, i$ respectively, with angle $r$, which don't include the endpoints. The intersection of these loci are precisely where the rays intersect, but as they are parallel rays, they will only intersect when one ray contains the other's endpoint, and in that case, the intersection is the entirety of the smaller ray.

This means we only need to consider two situations where the intersection is not empty:

• The ray starting at $-1$ includes $i$. This describes the ray starting at $i$ (i.e. $(x,y)=(0,1)$) with direction $(1,1)$, so $y=x+1$ and $x > 0$.
• The ray starting at $i$ includes $-1$. This describes the ray starting at $-1$ (i.e. $(x,y)=(-1,0)$) with direction $(-1,-1)$, so $y=x+1$ and $x < -1$.

The union of these gives the locus: $$y = x+1, \qquad x \not\in [-1,0]$$

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If the above seems too informal, it may be helpful to see the solution a couple other users have hinted about. Specifically, $z$ is on the locus if and only if $$\operatorname{arg}\left(\frac{z+1}{z-i}\right) = \operatorname{arg}(z+1) - \operatorname{arg}(z-i) = 0$$ which holds if and only if $\frac{z+1}{z-i} \in \mathbb{R}^+$, which holds if and only if $(z+1)\overline{(z-i)} \in \mathbb{R}^+$.

At this point, if we write $z = x+yi$, this becomes $$((x+1)+yi)(x+(1-y)i) = (x+1)x + y(y-1) + (x-y+1)i \in \mathbb{R}^+$$ or $$(x+1)x + y(y-1) > 0 \\ x-y+1 = 0$$ and solving this gives the same result we found above.

Answer 2

If $A(z_{A}), B(z_B), M(z)$ are three distinct points in the Argand plane, then $$\operatorname{arg}\frac{z_A-z}{z_B-z}$$ is a measure of the oriented angle of vectors $\angle(\vec{MB}, \vec{MA}).$

Set $z_A=-1, z_B=i.$ As $$\operatorname{arg}(z+1)=\operatorname{arg}(z-i)\quad \iff \quad \operatorname{arg}\frac{z+1}{z-i}=0\quad \iff \quad \operatorname{arg}\frac{-1-z}{i-z}=0,$$ the points $M$ are collinear with $A$ and $B,$ but they lie outside the segment $AB.$

Locus of points $M(z)$ satisfying the equation is the line $\overline{AB}$ excluding the segment $AB.$

Answer 3

We consider the three points $z,z+1$ and $z-i$.

• The line segment $[z,z+1]$ has length $1$ and is parallel to the $x$-axis.

• The line segment $[z,z-i]$ has length $1$ and is parallel to the $y$-axis.

• $\mathrm{arg}(z+1)=\mathrm{arg}(z-i)$ implies that the segment $[z-i,z+1]$ is part of the main-diagonal $x=y$, but not crossing the origin since otherwise $z-i$ and $z+1$ differ by an angle with $\pi$ radian. This is assured if $x\not\in[-1,0]$.

We conclude all solutions $z$ form with $z-i$ and $z+1$ an isosceles, right-angled triangle with hypotenuse $[z-i,z+1]$ on the main-diagonal and \begin{align*} \color{blue}{z=x+i(x+1)\qquad x\in\mathbb{R}\setminus[-1, 0]} \end{align*} being an element of the line parallel to the main-diagonal going through $i=(0,1)$.