Discussion
Let \(z\) be a complex number such that \(z^{3} + iz^{2} - iz + 1 = 0\) where \(i^{2} = -1\). Then |z| =
Answer: C
The value is |z| = 1.
Group the terms to factorise. Since \(-i \cdot i = 1\), the last two terms are \(-i(z + i)\).
\[ \begin{aligned} z^{3} + iz^{2} - iz + 1 &= z^{2}(z + i) - i(z + i) \\ &= (z + i)(z^{2} - i) \end{aligned} \]
So either z = -i, with |z| = 1, or \(z^{2} = i\), which gives \(|z|^{2} = |i| = 1\) and again |z| = 1. Every root lies on the unit circle, so 2, 3, 1/2 and 1/4 are impossible.
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