Equal distances and a square on the unit circle

Problem

On the complex plane, a complex number $z$ in quadrant I lies on the unit circle. There is also a fixed point

The distance from $w$ to $z$ equals the distance from $w$ to $z^3$. Find the real and imaginary parts of $z=a+bi$. The exact values and the answer format are on the CEEC 111 AST Mathematics I paper.

Where it goes wrong

The official item analysis records a score rate of $16\%$. The step that sticks is seeing $z^3$ and rushing into a long algebraic expansion, while forgetting that a complex absolute value is a distance on the complex plane. Redraw “equal distances” as the perpendicular bisector of a chord, and the cube only triples the argument.

Idea

Since $|z|=|z^3|=1$, the origin is also equally far from $z$ and $z^3$. So both the origin and $w$ lie on the perpendicular bisector of the chord $z{z}^3$.

Let the argument of $z$ be $\theta$. By De Moivre’s theorem, $z^3$ has argument $3\theta$. The midpoint of the chord points in direction $2\theta$, the direction of $z^2$. Because $z$ is in quadrant I, $z^2$ has argument $2\theta \in (0,\pi )$ and lies in the upper half-plane, and so does $w$. So the direction cannot flip to the opposite side, and since $|w|=1=|z^2|$,

Write $z=a+bi$, compare real and imaginary parts, and use $a^2+b^2=1$:

Adding the first and last equations gives $a^2=\frac{1}{5}$, and then $b^2=\frac{4}{5}$. Because $z$ is in quadrant I,

What the figure shows

• Drag Argument $\theta$ and watch $z^3$ move around the unit circle at three times the argument. The chord and its perpendicular bisector move with it.
• Compare the gold and blue segments from the fixed point $w$ to $z$ and to $z^3$. The sidebar shows the current difference between the two distances.
• Press Back to the solution position to check that the two distances are equal when $w=z^2$.