Question

Let $z$ be a complex number such that $z^{3}+iz^{2}-iz+1=0$ where $i^{2}=-1$. Then $|z|=$

• A. 2
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{1}{4}$
• E. 3

Hint:

$|z^n| = |z|^n$ is a useful property for complex equations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given a cubic equation in the complex variable z. We need to find the modulus of the solutions for z. A good approach is to try and factor the polynomial.
Step 2: Key Formula or Approach:
We will use factorization by grouping to simplify the equation \( z^3 + iz^2 - iz + 1 = 0 \).
We also know that \( 1 = -i^2 \). Substituting this can sometimes reveal factors.
Let's try substituting \( 1 = -i^2 \) for the last term.
\[ z^3 + iz^2 - iz - i^2 = 0 \] Step 3: Detailed Explanation:
Let's group the terms of the polynomial:
\[ (z^3 + iz^2) + (-iz - i^2) = 0 \] Factor out the common term from each group:
\[ z^2(z + i) - i(z + i) = 0 \] Now, factor out the common binomial term \( (z + i) \):
\[ (z^2 - i)(z + i) = 0 \] This gives two possibilities for the solutions:
Case 1: \( z + i = 0 \)
This implies \( z = -i \).
The modulus of z is \( |z| = |-i| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1 \).
Case 2: \( z^2 - i = 0 \)
This implies \( z^2 = i \).
To find \( |z| \), we can take the modulus of both sides of the equation:
\[ |z^2| = |i| \] Using the property \( |a^n| = |a|^n \), we get:
\[ |z|^2 = |i| \] The modulus of \( i \) is \( |0 + 1i| = \sqrt{0^2 + 1^2} = 1 \).
\[ |z|^2 = 1 \] Taking the square root of both sides, and knowing that modulus must be non-negative, we get:
\[ |z| = 1 \] Step 4: Final Answer:
In both possible cases, the modulus of z is 1. Therefore, \( |z| = 1 \).