The equation `Z^(3)+iZ-1=0` has

Options

A. three real roots
B. one real roots
C. no real roots
D. no real or complex roots

Explanation

To solve the equation \( Z^3 + iZ - 1 = 0 \) and determine the nature of its roots, we will follow these steps:

### Step 1: Substitute \( Z \) with \( x \)
We start by rewriting the equation:
\[ x^3 + ix - 1 = 0 \]

### Step 2: Analyze the equation
We can separate the real and imaginary parts of the equation. The real part is \( x^3 - 1 \) and the imaginary part is \( ix \). For the equation to hold, both parts must equal zero.

### Step 3: Set up the equations
From the real part:
\[ x^3 - 1 = 0 \]
From the imaginary part:
\[ ix = 0 \implies x = 0 \]

### Step 4: Solve the real part
The equation \( x^3 - 1 = 0 \) can be factored as:
\[ x^3 = 1 \]
This gives us the roots:
\[ x = 1 \]
The roots of \( x^3 = 1 \) are \( x = 1, \omega, \omega^2 \) where \( \omega = e^{2\pi i / 3} \) (the complex cube roots of unity).

### Step 5: Check for real roots
We already found \( x = 0 \) from the imaginary part. Now we check if \( x = 1 \) satisfies the original equation:
\[ 1^3 + i(1) - 1 = 1 + i - 1 = i \neq 0 \]
Thus, \( x = 1 \) is not a root.

### Step 6: Conclusion about the roots
Since \( x = 0 \) does not satisfy the equation and \( x = 1 \) also does not satisfy it, we conclude that there are no real roots. The complex roots are \( \omega \) and \( \omega^2 \).

### Final Answer
The equation \( Z^3 + iZ - 1 = 0 \) has **no real roots**.

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