The complex number z = a + bi satisfies (a + bi)^2025 = (a - bi)^2025. What does this condition mean about z?

Step-by-step solution

1. 1

   Rewrite the right-hand side as a conjugate. Since $a-bi=\overline{z}$ and conjugation commutes with powers, $\overline{z}^{\,2025}=\overline{z^{2025}}$. The hypothesis therefore says

   $z^{2025}=\overline{z^{2025}}.$

   That single rewrite is the whole idea: the equation is not really about $z$, it is about the number $w=z^{2025}$.

2. 2

   Use the fact that $w=\overline{w}$ characterises real numbers. Writing $w=u+vi$, the equation $w=\overline{w}$ gives $u+vi=u-vi$, so $2vi=0$ and $v=0$. Hence the condition is exactly

   $z^{2025}\in\mathbb{R}.$

   So the plain-language meaning is: the 2025th power of $a+bi$ is a real number.

3. 3

   Translate into polar form to see which $z$ qualify. Write $z=re^{i\theta}$ with $r=\sqrt{a^{2}+b^{2}}$. Then $\overline{z}=re^{-i\theta}$, and by de Moivre

   $z^{2025}=r^{2025}e^{i\,2025\theta},\qquad \overline{z}^{\,2025}=r^{2025}e^{-i\,2025\theta}.$

   If $r=0$ — that is $a=b=0$ — both sides are $0$ and the condition holds trivially.

4. 4

   Solve the angle equation for $z\neq 0$. Cancelling $r^{2025}\neq 0$ leaves $e^{i2025\theta}=e^{-i2025\theta}$, i.e. $e^{i4050\theta}=1$, so

   $4050\theta=2k\pi\;\Longrightarrow\;\theta=\frac{k\pi}{2025},\qquad k\in\mathbb{Z}.$

   Equivalently $\sin(2025\theta)=0$, which is the same statement that the imaginary part of $z^{2025}$ vanishes. Notice the modulus $r$ is completely unconstrained — only the direction of $z$ matters.

5. 5

   Describe the solution set geometrically. The admissible $z$ form $2025$ lines through the origin (each line covering two of the $4050$ angles $k\pi/2025$), spaced $\pi/2025=180^{\circ}/2025$ apart, plus the origin itself. The real axis ($k=0$) and the purely imaginary case are included only when $\theta$ is a multiple of $\pi/2025$ — and since $2025$ is odd, $\theta=\pi/2$ is not of that form, so $z=i$ does not satisfy the condition. Indeed $i^{2025}=i$, while $(-i)^{2025}=-i$ ✓.