Solving Complex Numbers Equation: Finding Roots and Their Sum

The problem asks for the sum of three non-zero complex numbers, $z_1, z_2, z_3$, which all satisfy the equation $z^2 = i\bar{z}$. This means that $z_1, z_2, z_3$ are the distinct non-zero roots of the equation $z^2 = i\bar{z}$. To find their sum, we first need to find these roots.

Finding the Roots of $z^2 = i\bar{z}$

We can solve this equation by representing the complex number $z$ in its polar form. Let $z = re^{i\theta}$, where $r$ is the magnitude ($r \ge 0$) and $\theta$ is the argument. The conjugate of $z$ is $\bar{z} = re^{-i\theta}$. Also, $i$ can be written in polar form as $i = 1 \cdot e^{i\pi/2}$.

Substitute these forms into the equation $z^2 = i\bar{z}$:

Since $z$ is a non-zero complex number, its magnitude $r$ must be greater than 0 ($r > 0$). We can divide both sides by $r$:

For two complex numbers in polar form to be equal, their magnitudes must be equal, and their arguments must be equal up to a multiple of $2\pi$.

• Equating magnitudes: $r = 1$.
• Equating arguments: $2\theta = \frac{\pi}{2} - \theta + 2k\pi$, where $k$ is an integer ($k \in \mathbb{Z}$).

Now, we solve for $\theta$ from the argument equation:

We need to find distinct values of $\theta$ for integer values of $k$. The argument $\theta$ is typically considered within the interval $(-\pi, \pi]$ or $[0, 2\pi)$. Let's use $[0, 2\pi)$.

• For $k=0$: $\theta_1 = \frac{\pi}{6}$. This gives the root $z_1 = 1 \cdot e^{i\pi/6} = \cos(\pi/6) + i\sin(\pi/6) = \frac{\sqrt{3}}{2} + \frac{1}{2}i$.
• For $k=1$: $\theta_2 = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{6} + \frac{4\pi}{6} = \frac{5\pi}{6}$. This gives the root $z_2 = 1 \cdot e^{i5\pi/6} = \cos(5\pi/6) + i\sin(5\pi/6) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$.
• For $k=2$: $\theta_3 = \frac{\pi}{6} + \frac{4\pi}{3} = \frac{\pi}{6} + \frac{8\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}$. This gives the root $z_3 = 1 \cdot e^{i3\pi/2} = \cos(3\pi/2) + i\sin(3\pi/2) = 0 - i = -i$.
• For $k=3$: $\theta_4 = \frac{\pi}{6} + \frac{6\pi}{3} = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}$, which is equivalent to $\frac{\pi}{6}$ (since $\frac{13\pi}{6} - 2\pi = \frac{\pi}{6}$). The roots start repeating.

Thus, there are three distinct non-zero roots satisfying the equation $z^2 = i\bar{z}$. These are the complex numbers $z_1, z_2, z_3$.

The three roots are:

• $z_1 = \frac{\sqrt{3}}{2} + \frac{1}{2}i$
• $z_2 = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$
• $z_3 = -i$

Calculating the Sum $z_1 + z_2 + z_3$

Now we need to find the sum of these three complex numbers:

Group the real parts and the imaginary parts:

So, the sum is:

The sum of the three non-zero complex numbers satisfying $z^2 = i\bar{z}$ is 0.

Revision Table: Complex Numbers & Equations

The equation $z^2 = i\bar{z}$ can be viewed as a type of polynomial-like equation involving both $z$ and $\bar{z}$. By converting to polar form, we simplify the equation into separate conditions for magnitude and argument, which is a common technique for solving complex number equations.

Additional Information: Geometric Interpretation of Roots

The roots we found have a geometric interpretation. All roots have a magnitude of $r=1$, meaning they lie on the unit circle in the complex plane. Their arguments are $\pi/6$, $5\pi/6$, and $3\pi/2$.

Let's visualize these points on the complex plane:

• $z_1$: Angle $\pi/6$ (30 degrees) from the positive real axis.
• $z_2$: Angle $5\pi/6$ (150 degrees) from the positive real axis.
• $z_3$: Angle $3\pi/2$ (270 degrees) from the positive real axis, which is along the negative imaginary axis.

These three points form a triangle inscribed in the unit circle. The fact that their sum is zero relates to properties of the roots of certain polynomial equations, specifically when the roots are symmetrically distributed around the origin, although the original equation $z^2 = i\bar{z}$ is not a simple polynomial in $z$ alone.

This problem effectively tests the ability to work with complex numbers in polar form and solve equations involving conjugates.