Question

If \( x^6 = (\sqrt{3}-i)^5 \), then the product of all of its roots is

• A. \( 2^5(\sqrt{3}+i) \)
• B. \( \frac{2^6}{\sqrt{3}+i} \)
• C. \( 2^6(\sqrt{3}-i) \)
• D. \( \frac{2^6}{\sqrt{3}-i} \)

Hint:

For \(x^n = c\), i.e., \(x^n - c = 0\), the product of roots is \( (-1)^n (-c/1) = (-1)^{n+1}c \). Here \(n=6\), so product is \((-1)^7 c = -c\). Let \(c = (\sqrt{3}-i)^5\). Convert \(\sqrt{3}-i\) to polar form: \(2e^{-i\pi/6}\). So, product \( = -(2e^{-i\pi/6})^5 = -32e^{-i5\pi/6}\). Using \(-1 = e^{i\pi}\), product \( = 32e^{i\pi}e^{-i5\pi/6} = 32e^{i\pi/6} = 32(\frac{\sqrt{3}}{2} + \frac{i}{2}) = 16(\sqrt{3}+i)\). Option (4) simplifies to \(16(\sqrt{3}+i)\).

Answer 1

The Correct Option is D

The given equation is \( x^6 = (\sqrt{3}-i)^5 \).
Let \(C = (\sqrt{3}-i)^5\).
The equation is \(x^6 - C = 0\).
For a polynomial equation \(a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 = 0\), the product of the roots is \(P = (-1)^n \frac{a_0}{a_n}\).
In this case, \(n=6\), \(a_6=1\), and the constant term \(a_0 = -C\).
So, the product of the roots is \(P = (-1)^6 \frac{-C}{1} = (1)(-C) = -C\).
Thus, \(P = -(\sqrt{3}-i)^5\).
Let \(w = \sqrt{3}-i\).
Convert \(w\) to polar form \(r(\cos\theta + i\sin\theta)\).
\(r = |w| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2\).
For the argument \(\theta\), \(\cos\theta = \frac{\sqrt{3}}{2}\) and \(\sin\theta = \frac{-1}{2}\).
This places \(\theta\) in the fourth quadrant, so \(\theta = -\frac{\pi}{6}\).
So, \(w = 2\left(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})\right) = 2e^{-i\pi/6}\).
Then \(C = w^5 = \left(2e^{-i\pi/6}\right)^5 = 2^5 e^{-i5\pi/6}\).
The product of roots is \(P = -C = -2^5 e^{-i5\pi/6}\).
We can write \(-1\) in polar form as \(e^{i\pi}\).
So, \(P = e^{i\pi} \cdot 2^5 e^{-i5\pi/6} = 2^5 e^{i(\pi - 5\pi/6)} = 2^5 e^{i\pi/6}\).
Convert \(P\) back to rectangular form: \(P = 2^5 \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 32 \left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = 16(\sqrt{3}+i)\).
Now, simplify option (4): \( \frac{2^6}{\sqrt{3}-i} \).
\( \frac{2^6}{\sqrt{3}-i} = \frac{64}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} = \frac{64(\sqrt{3}+i)}{(\sqrt{3})^2 - (i)^2} = \frac{64(\sqrt{3}+i)}{3 - (-1)} = \frac{64(\sqrt{3}+i)}{4} = 16(\sqrt{3}+i) \).
This matches the calculated product \(P\).
\[ \boxed{\frac{2^6}{\sqrt{3}-i}} \]