Question:

If \[ (1+\sin\alpha+i\cos\alpha)^8 = A(\cos\theta+i\sin\theta), \] then $A$ and $\theta$ are respectively

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When a complex number is raised to a power, use De Moivre's theorem after converting it into polar form.
Updated On: Jun 3, 2026
  • \[ 2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),\;2\alpha \]
  • \[ 2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),\;-4\alpha \]
  • \[ 2^8\cos^8\left(\frac{\pi}{4}+\frac{\alpha}{2}\right),\;2\alpha \]
  • \[ 2^8\cos^8\left(\frac{\pi}{4}+\frac{\alpha}{2}\right),\;-4\alpha \]
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The Correct Option is B

Solution and Explanation

Step 1: Concept
Convert the complex number into polar form.

Step 2: Meaning
Observe that \[ 1+\sin\alpha+i\cos\alpha = (1+\sin\alpha)+i\cos\alpha. \]

Step 3: Analysis
Its modulus is \[ r=\sqrt{(1+\sin\alpha)^2+\cos^2\alpha} =\sqrt{2(1+\sin\alpha)} =2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right). \] Its argument is \[ \phi=\frac{\pi}{4}+\frac{\alpha}{2}. \] Hence \[ 1+\sin\alpha+i\cos\alpha = 2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \left(\cos\phi+i\sin\phi\right). \] Applying De Moivre's theorem, \[ A= 2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), \] and the argument becomes \[ 8\phi = 2\pi-4\alpha. \] Therefore the principal argument is \[ -4\alpha. \]

Step 4: Conclusion
Thus \[ A= 2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), \qquad \theta=-4\alpha. \]

Final Answer: (B)
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