Question

The solutions of the equation \[ 2\sqrt{2}\,x^4=(\sqrt{3}-1)+i(\sqrt{3}+1) \] are

• A. \[ x=\pm \operatorname{cis}\frac{3\pi}{38}, \quad \pm \operatorname{cis}\frac{23\pi}{38} \]
• B. \[ x=\pm \operatorname{cis}\frac{5\pi}{48}, \quad \pm \operatorname{cis}\frac{29\pi}{48} \]
• C. \[ x=\pm \operatorname{cis}\frac{7\pi}{48}, \quad \pm \operatorname{cis}\frac{41\pi}{48} \]
• D. \[ x=\pm \operatorname{cis}\frac{9\pi}{62}, \quad \pm \operatorname{cis}\frac{27\pi}{62} \] \bigskip

Hint:

For equations of the form \[ x^n=r\,\operatorname{cis}\theta, \] the \(n\) roots are \[ x=r^{1/n} \operatorname{cis} \left( \frac{\theta+2k\pi}{n} \right), \quad k=0,1,\ldots,n-1. \] Always convert the complex number to polar form before applying De Moivre's theorem.

Answer 1

The Correct Option is B

Concept: Express the complex number in trigonometric form and apply De Moivre's theorem to obtain all fourth roots.

Step 1: The modulus of \((\sqrt3-1)+i(\sqrt3+1)\) is \(2\sqrt2\) and its argument is \(\frac{5\pi}{12}\). Hence \(2\sqrt2\,x^4=2\sqrt2\,\operatorname{cis}\frac{5\pi}{12}\), so \(x^4=\operatorname{cis}\frac{5\pi}{12}\).

Step 2: Therefore \(x=\operatorname{cis}\left(\frac{5\pi}{48}+\frac{k\pi}{2}\right),\;k=0,1,2,3\).

Step 3: Hence the four roots are \(\boxed{x=\pm\operatorname{cis}\frac{5\pi}{48},\;\pm\operatorname{cis}\frac{29\pi}{48}}\).