Find all non-zero complex numbers Z satisfying Z̄ = iZ².

Q: Wait, why can we just divide by $|Z|$? What if $|Z|=0$?

That is a great observation! The problem explicitly asks for non-zero complex numbers, so we are given $Z \neq 0$. This means $|Z|$ cannot be zero, allowing us to safely divide by it. If $Z$ were zero, the equation would be $0=0$, which is true, but we are looking for the other, more interesting solutions!

Q: Why do we add $2k\pi$ to the exponent?

Remember that the complex exponential function $e^{i\theta}$ is periodic with a period of $2\pi$. This means $e^{i\theta} = e^{i(\theta + 2k\pi)}$ for any integer $k$. When we equate the exponents of two complex exponentials, we must account for all possible angles that represent the same point on the complex plane.

Q: How do we know when to stop picking values for $k$?

Excellent question. Since we are solving for $\theta$ in the expression $\theta = -\frac{\pi}{6} - \frac{2k\pi}{3}$, we only need to find values of $k$ that produce unique angles within the range $[0, 2\pi)$ or $(-\pi, \pi]$. For $k=0, 1, 2$, we get distinct angles. If you pick $k=3$, you will just get an angle that is $2\pi$ greater than the one for $k=0$, which represents the same complex number. Thus, we only have three distinct solutions.

Question

Find all non-zero complex numbers $Z$ satisfying $\bar{Z} = iZ^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with solving the equation $\bar{Z} = iZ^2$ for all non-zero complex numbers $Z$. This equation represents a fascinating intersection of algebraic manipulation and geometric rotation within the complex plane.

### The Modulus Insight

When dealing with equations involving both $Z$ and its conjugate $\bar{Z}$, the modulus is our most effective tool. By taking the modulus of both sides, we isolate the distance of the solutions from the origin:

$$|\bar{Z}| = |iZ^2|$$

Using the properties of the modulus, where $|\bar{Z}| = |Z|$ and $|iZ^2| = |i||Z|^2$, and knowing that $|i| = 1$, the equation simplifies to:

$$|Z| = |Z|^2$$

Since we are searching for non-zero solutions, we divide by $|Z|$ to find $|Z| = 1$. Geometrically, this confirms that all solutions must lie on the **unit circle**.

### The Euler Transformation

With the constraint $|Z| = 1$, we can represent any solution using Euler's formula, $Z = e^{i	heta}$. Substituting this into the original equation, we have $\bar{Z} = e^{-i	heta}$ and $Z^2 = e^{i2	heta}$.

The equation becomes:

$$e^{-i	heta} = i e^{i2	heta}$$

To solve for $	heta$, we express the imaginary unit $i$ as $e^{i\pi/2}$. The equation then transforms into:

$$e^{-i	heta} = e^{i\pi/2} \cdot e^{i2	heta} = e^{i(\pi/2 + 2	heta)}$$

### The Periodic Dance

Because the complex exponential is periodic, we equate the exponents by accounting for the $2k\pi$ rotation factor, where $k$ is an integer:

$$-	heta = \frac{\pi}{2} + 2	heta + 2k\pi$$

Rearranging the terms to isolate $	heta$ yields:

$$-3	heta = \frac{\pi}{2} + 2k\pi$$

$$	heta = -\frac{\pi}{6} - \frac{2k\pi}{3}$$

### Final Calculation

We now evaluate the expression for $k = 0, 1, 2$ to find the distinct roots:

For $k=0$:
$$	heta = -\frac{\pi}{6} \Rightarrow Z_1 = \cos\left(-\frac{\pi}{6}ight) + i\sin\left(-\frac{\pi}{6}ight) = \frac{\sqrt{3}}{2} - \frac{i}{2}$$

For $k=1$:
$$	heta = -\frac{\pi}{6} - \frac{2\pi}{3} = -\frac{5\pi}{6} \Rightarrow Z_2 = \cos\left(-\frac{5\pi}{6}ight) + i\sin\left(-\frac{5\pi}{6}ight) = -\frac{\sqrt{3}}{2} - \frac{i}{2}$$

For $k=2$:
$$	heta = -\frac{\pi}{6} - \frac{4\pi}{3} = -\frac{3\pi}{2} \equiv \frac{\pi}{2} \Rightarrow Z_3 = \cos\left(\frac{\pi}{2}ight) + i\sin\left(\frac{\pi}{2}ight) = i$$

The set of solutions is **$\{i, \frac{\sqrt{3}}{2} - \frac{i}{2}, -\frac{\sqrt{3}}{2} - \frac{i}{2}\}$**. These points form a perfect equilateral triangle inscribed within the unit circle.

Animated Solution for Mathematics - Complex Numbers: Find all non-zero complex numbers $Z$ satisfying $\overset{ˉ}{Z} = i Z^{2}$ .

Visualized Solution

Visualizing the Equation $\overset{ˉ}{Z} = i Z^{2}$

Taking the Modulus

Simplifying the Magnitude

The Unit Circle Constraint

Using Euler's Form $Z = e^{i θ}$

Substituting into the Equation

Converting $i$ to Exponential Form

Equating the Exponents

Solving for $θ$

Finding the First Root ( $k = 0$ )

Finding the Second Root ( $k = 1$ )

Finding the Third Root ( $k = 2$ )

Summary and Final Takeaway

The Sigma Insight: Euler's Form and De Moivre's Theorem

Analyzing the Setup

The Modulus Insight

The Euler Transformation

The Periodic Dance

Final Calculation

Similar Questions

Let $\overset{z}{ˉ}$ denote the complex conjugate of a complex number $z$ . If $z$ is a non-zero complex number for which both real and imaginary parts of $(\overset{z}{ˉ})^{2} + \frac{1}{z ^{2}}$ are integers, then which of the following is/are possible value(s) of $∣ z ∣$ ?

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find all non-zero complex numbers Z satisfying Zˉ=iZ2.

Visualized Solution

Visualizing the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zˉ=iZ2

Taking the Modulus

Simplifying the Magnitude

The Unit Circle Constraint

Using Euler's Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z=eiθ

Substituting into the Equation

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i to Exponential Form

Equating the Exponents

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Finding the First Root ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=0)

Finding the Second Root ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=1)

Finding the Third Root ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=2)