Find z Given Arg(z + 1) = π/6 and Arg(z − 1) = 2π/3

Assume that $\operatorname{Arg}$ means the principal argument, measured counterclockwise from the positive real axis. The condition $\operatorname{Arg}(z+1)=\pi/6$ places $z+1$ on a ray at angle $\pi/6$. The second condition places $z-1$ on a ray at angle $2\pi/3$. Since these expressions shift the point by $1$ in opposite directions, the two rays begin at $-1$ and $1$ on the real axis.

Their intersection is

$$z=\frac{1}{2}+\frac{\sqrt{3}}{2}i.$$

Turn each argument into a ray

For a nonzero complex number $w$, an equation $\operatorname{Arg}(w)=\theta$ means that $w$ lies on the ray making angle $\theta$ with the positive real axis. In polar form, $w=r(\cos\theta+i\sin\theta)$ with $r>0$. The strict positivity matters: the origin has no defined argument.

Apply this to each expression:

$$z+1=r\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right),\qquad r>0,$$ $$z-1=s\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right),\qquad s>0.$$

Here $r$ and $s$ are distances along the rays, not necessarily the same distance.

Solve the intersection

Using $\cos(\pi/6)=\sqrt{3}/2$, $\sin(\pi/6)=1/2$, $\cos(2\pi/3)=-1/2$, and $\sin(2\pi/3)=\sqrt{3}/2$, write both expressions for $z$:

$$z=-1+\frac{\sqrt{3}}{2}r+\frac{r}{2}i,$$ $$z=1-\frac{s}{2}+\frac{\sqrt{3}}{2}s i.$$

Equate the real and imaginary parts. The imaginary parts give

$$\frac{r}{2}=\frac{\sqrt{3}}{2}s,$$

so $r=\sqrt{3}s$. The real parts give

$$-1+\frac{\sqrt{3}}{2}r=1-\frac{s}{2}.$$

Substitute $r=\sqrt{3}s$:

$$-1+\frac{3s}{2}=1-\frac{s}{2}, \qquad 2s=2, \qquad s=1.$$

Thus $r=\sqrt{3}$. Substituting into either expression for $z$ yields

$$z=1-\frac{1}{2}+\frac{\sqrt{3}}{2}i =\frac{1}{2}+\frac{\sqrt{3}}{2}i.$$

Verify both arguments

For the proposed point,

$$z+1=\frac{3}{2}+\frac{\sqrt{3}}{2}i.$$

Its real and imaginary parts are positive, and their ratio is $1/\sqrt{3}$, so its principal argument is $\pi/6$. Also,

$$z-1=-\frac{1}{2}+\frac{\sqrt{3}}{2}i.$$

This lies in the second quadrant, and its reference angle is $\pi/3$, so its principal argument is $\pi-\pi/3=2\pi/3$. Both conditions hold.

Geometric interpretation

In the complex plane, $z$ is the intersection of a ray starting at $-1$ and rising at $30^\circ$, and a ray starting at $1$ and rising at $120^\circ$. The ray directions are not parallel, so their supporting lines meet at one point. The calculation above also confirms that the intersection lies forward along both rays, because the distances are $r=\sqrt{3}>0$ and $s=1>0$.

You can also use slopes to locate the supporting-line intersection. Writing $z=x+iy$, the two lines are

$$y=\frac{x+1}{\sqrt{3}},\qquad y=-\sqrt{3}(x-1).$$

Solving gives $x=1/2$ and $y=\sqrt{3}/2$. When using this shorter method, check that the point is on the rays, not just on the full lines; a line intersection can fall behind one of the ray origins.

Common mistakes

• Treating the argument as a length. The angle determines a direction; the positive parameters $r$ and $s$ determine how far to travel on each ray.
• Using the wrong starting point. $z+1$ measures from $-1$, while $z-1$ measures from $1$.
• Ignoring quadrants. A slope alone does not determine an angle. In particular, a point with negative real part and positive imaginary part is in the second quadrant, as required for $2\pi/3$.
• Allowing zero distance. Neither $z=-1$ nor $z=1$ is valid, because one of the arguments would be undefined.

If lowercase $\arg$ is being used as a multivalued argument rather than $\operatorname{Arg}$ as the principal argument, the same directions are understood modulo $2\pi$. The intersection and the solution remain unchanged.