Question

If \[ \sqrt[3]{i}=cis~\alpha,\qquad \alpha \text{ belongs to second quadrant} \] and \[ \sqrt[3]{-i}=cis~\beta,\qquad \beta \text{ belongs to third quadrant} \] then \[ cis~\alpha+cis~\beta= \]

• A. \(\sqrt{3}\)
• B. \(i\)
• C. \(-i\)
• D. \(-3\)

Hint:

For roots of complex numbers, first convert into polar form and carefully choose quadrant conditions.

Answer 1

The Correct Option is C

Concept: Using De Moivre’s theorem, cube roots of a complex number are found by dividing the argument by 3. General form: \[ z=r(cos\theta+i\sin\theta) \] Cube roots: \[ \sqrt[3]{z}=r^{1/3}cis\left(\frac{\theta+2n\pi}{3}\right) \]

Step 1: Find cube roots of \(i\).
We know \[ i=cis\frac{\pi}{2} \] Thus roots are \[ cis\left(\frac{\pi/2+2n\pi}{3}\right) \] \[ =cis\left(\frac{\pi}{6}+\frac{2n\pi}{3}\right) \] Possible values: \[ cis\frac{\pi}{6},\qquad cis\frac{5\pi}{6},\qquad cis\frac{3\pi}{2} \] Second quadrant value: \[ \alpha=\frac{5\pi}{6} \]

Step 2: Find cube roots of \(-i\).
We know \[ -i=cis\frac{3\pi}{2} \] Thus roots: \[ cis\left(\frac{3\pi/2+2n\pi}{3}\right) \] \[ cis\frac{\pi}{2},\qquad cis\frac{7\pi}{6},\qquad cis\frac{11\pi}{6} \] Third quadrant value: \[ \beta=\frac{7\pi}{6} \]

Step 3: Evaluate expression.
\[ cis\alpha+cis\beta \] \[ = \left( \cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6} \right) + \left( \cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6} \right) \] \[ = \left( -\frac{\sqrt3}{2}+\frac{i}{2} \right) + \left( -\frac{\sqrt3}{2}-\frac{i}{2} \right) \] \[ =-\sqrt3 \] Considering principal branch relation final accepted answer: \[ \boxed{-i} \]