Question

If z is a complex number with modulus 1, then the equation $\left( \frac{1 + ia}{1 - ia} \right)^{4}$ = z has –

• A. All roots real and distinct
• B. Two roots real and two roots imaginary
• C. Three roots real and one root imaginary
• D. One root real and three roots imaginary

Answer 1

Answer: (A)

All roots real and distinct

Answer 2

Sol. We have $\left( \frac{1 + ia}{1 - ia} \right)^{4}$= z and |z| = 1.

Let $\left( \frac{1 + ia}{1 - ia} \right)^{4}$= cos q + i sin q

Ž $\frac{1 + ia}{1 - ia}$ = (cos q + i sin q)^1/4

= (cos (2kp + q) + sin (2kp + q))^1/4 , k Ī Z

= cos $\left( \frac{2k\pi + \theta}{4} \right)$ + i sin $\left( \frac{2k\pi + \theta}{4} \right)$, k = 0, 1, 2, 3

\ $\frac{1 + ia}{1 - ia}$ = cos a + i sin a, where a = $\frac{2k\pi + \theta}{4}$

Ž $\frac{2}{2ia}$= $\frac{\cos\alpha + i\sin\alpha + 1}{\cos\alpha + i\sin\alpha - 1}$

=$\frac{2\cos^{2}\frac{\alpha}{2} + 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{- 2\sin^{2}\frac{\alpha}{2} + 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}$ = $\frac{2\cos\frac{\alpha}{2}\left( \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2} \right)}{2i\sin\frac{\alpha}{2}\left( \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2} \right)}$

= $\frac{1}{i}$cot $\frac{\alpha}{2}$

\ ia = i tan $\frac{\alpha}{2}$Ž a = tan $\frac{\alpha}{2}$

\ a = tan$\frac{2k\pi + \theta}{8}$, k = 0, 1, 2, 3 = tan$\frac{\theta}{8}$, tan $\left( \frac{\pi}{4} + \frac{\theta}{8} \right)$, tan$\left( \frac{\pi}{2} + \frac{\theta}{8} \right)$, tan$\left( \frac{3\pi}{4} + \frac{\theta}{8} \right)$

\ All roots are real and distinct.