Let's solve the equation for complex numbers \( z \): \[ z^5 - \overline{z} = 0 \] **Step 1: Express \(z\) in polar form** Let \( z = re^{i\theta} \), where \( r \ge 0 \), \( \theta \in [0, 2\pi) \). The conjugate is: \[ \overline{z} = r e^{-i\theta} \] Plug into the equation: \[ (z)^5 = \overline{z} \] \[ (r e^{i\theta})^5 = r e^{-i\theta} \] \[ r^5 e^{i 5\theta} = r e^{-i\theta} \] If \( r = 0 \), then \( z = 0 \) is a solution. Let's check: \[ 0^5 - \overline{0} = 0 - 0 = 0 \] So \( z = 0 \) **is a solution**. Now, for \( r \ne 0 \), divide both sides by \( r \): \[ r^4 e^{i 5\theta} = e^{-i\theta} \] \[ r^4 = 1, \quad e^{i 5\theta} = e^{-i\theta} \] **Step 2: Solve for \( r \)** \[ r^4 = 1 \implies r = 1 \] (since \( r \) is the modulus and must be non-negative, and only \( r=1 \) makes sense for modulus.) **Step 3: Solve for \( \theta \)** \[ e^{i 5\theta} = e^{-i\theta} \implies 5\theta = -\theta + 2\pi k, \quad k\in\mathbb{Z} \] \[ 6\theta = 2\pi k \implies \theta = \frac{\pi k}{3} \] where \( k = 0, 1, 2, 3, 4, 5 \) (since \( \theta \) in \( [0, 2\pi) \), and with period \( 2\pi \), 6 distinct values). **Step 4: List all solutions** Thus, the solutions are: - \( z_0 = 0 \) (the origin) - \( z_k = e^{i\theta_k} \), where \( \theta_k = \frac{\pi k}{3} \), \( k = 0, 1, 2, 3, 4, 5 \) In **standard form** (\( z = x + iy \)), these are: \[ z_k = \cos\left(\frac{\pi k}{3}\right) + i\sin\left(\frac{\pi k}{3}\right) \] for \( k = 0, 1, 2, 3, 4, 5 \). Explicitly, the six unit roots at these angles are: | \(k\) | \( \theta_k \) | \( z_k \) in standard form | |---|---|---| | 0 | 0 | \( 1 + 0i \) | | 1 | \( \frac{\pi}{3} \) | \( \frac{1}{2} + i\frac{\sqrt{3}}{2} \) | | 2 | \( \frac{2\pi}{3} \) | \( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) | | 3 | \( \pi \) | \( -1 + 0i \) | | 4 | \( \frac{4\pi}{3} \) | \( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \) | | 5 | \( \frac{5\pi}{3} \) | \( \frac{1}{2} - i\frac{\sqrt{3}}{2} \) | **Step 5: Sketch on the complex plane** - The origin \( z=0 \) (center). - The other 6 solutions are the 6th roots of unity, equally spaced on the unit circle at angles \( 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \). Here is an ASCII sketch (the circle is the unit circle): ``` | Im | * | * | | --*---------O---------*-- Re | | * | * | ``` * `O` is the origin (solution \( z = 0 \)). * The `*` are the six nonzero solutions, evenly spaced around the unit circle. --- **Summary of solutions:** - **Polar form:** \( z = 0 \) and \( z = e^{i\frac{\pi k}{3}} \), \( k = 0,1,2,3,4,5 \) - **Standard form:** \( z = 0 \) and - \( z_0 = 1 \) - \( z_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2} \) - \( z_2 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) - \( z_3 = -1 \) - \( z_4 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \) - \( z_5 = \frac{1}{2} - i\frac{\sqrt{3}}{2} \) **All solutions are the origin and the 6th roots of unity.**