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AP EAMCET · Maths · Complex Number

If \(Z \neq 0\) is a complex number such that \(Z^2+Z|Z|+|Z|^2=0\) then \(Z\) is in the set (Here \(\omega\) is a complex cube root of unity)

  1. A \(\{1\}\)
  2. B \(\{i,-i\}\)
  3. C \(\left\{\omega, \omega^2\right\}\)
  4. D \(\phi\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left\{\omega, \omega^2\right\}\)

Step-by-step Solution

Detailed explanation

\(\frac{Z^2}{|Z|^2} + \frac{Z}{|Z|} + 1 = 0\) Let \(w = \frac{Z}{|Z|}\). Then \(w^2 + w + 1 = 0\). \(w = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\). \(w = \omega\) or \(w = \omega^2\). \(\frac{Z}{|Z|} = \omega\) or…