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: To solve equations involving complex numbers, first express the complex number in polar form \[ z=r(\cos\theta+i\sin\theta) =r\,\operatorname{cis}\theta. \] Then apply De Moivre's theorem: \[ z^{\frac1n} = r^{\frac1n} \operatorname{cis} \left( \frac{\theta+2k\pi}{n} \right), \qquad k=0,1,\ldots,n-1. \]

Step 1: Express the given complex number in polar form. Given \[ 2\sqrt2\,x^4 = (\sqrt3-1)+i(\sqrt3+1). \] Let \[ z=(\sqrt3-1)+i(\sqrt3+1). \] Its modulus is \[ |z| = \sqrt{(\sqrt3-1)^2+(\sqrt3+1)^2}. \] \[ = \sqrt{(4-2\sqrt3)+(4+2\sqrt3)} = \sqrt8 = 2\sqrt2. \] Therefore, \[ z = 2\sqrt2 \left( \cos\theta+i\sin\theta \right). \] Now, \[ \tan\theta = \frac{\sqrt3+1}{\sqrt3-1}. \] Rationalizing, \[ \tan\theta = \frac{(\sqrt3+1)^2}{3-1} = \frac{4+2\sqrt3}{2} = 2+\sqrt3. \] Since \[ \tan\frac{5\pi}{12} = 2+\sqrt3, \] we get \[ \theta=\frac{5\pi}{12}. \] Hence, \[ (\sqrt3-1)+i(\sqrt3+1) = 2\sqrt2\, \operatorname{cis}\frac{5\pi}{12}. \]

Step 2: Find \(x^4\). Substituting, \[ 2\sqrt2\,x^4 = 2\sqrt2\, \operatorname{cis}\frac{5\pi}{12}. \] Dividing by \(2\sqrt2\), \[ x^4 = \operatorname{cis}\frac{5\pi}{12}. \]

Step 3: Find the fourth roots. Using De Moivre's theorem, \[ x = \operatorname{cis} \left( \frac{\frac{5\pi}{12}+2k\pi}{4} \right), \qquad k=0,1,2,3. \] Thus, \[ x = \operatorname{cis} \left( \frac{5\pi}{48} +\frac{k\pi}{2} \right). \] For \(k=0,1,2,3\), \[ x= \operatorname{cis}\frac{5\pi}{48}, \] \[ x= \operatorname{cis}\frac{29\pi}{48}, \] \[ x= \operatorname{cis}\frac{53\pi}{48}, \] \[ x= \operatorname{cis}\frac{77\pi}{48}. \]

Step 4: Express the roots in the required form. Since \[ \operatorname{cis}\frac{53\pi}{48} = -\operatorname{cis}\frac{5\pi}{48}, \] and \[ \operatorname{cis}\frac{77\pi}{48} = -\operatorname{cis}\frac{29\pi}{48}, \] the four roots can be written as \[ x = \pm \operatorname{cis}\frac{5\pi}{48}, \qquad \pm \operatorname{cis}\frac{29\pi}{48}. \]

Step 5: Write the final answer. \[ \boxed{ x = \pm \operatorname{cis}\frac{5\pi}{48}, \quad \pm \operatorname{cis}\frac{29\pi}{48} } \]