The product of all the values of ({3} - i)³/7} is

Correct Answer is : −8i-8i−8i

Correct Answer is : −8i-8i−8i

Test Index

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 11 of 160

Marks: +1, -0

(\sqrt{3} - i)^{\frac{3}{7}}

Solution:

Let

z = \sqrt{3} - i

|z| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2

And

\theta = \tan^{-1} \left( \frac{-1}{\sqrt{3}} \right) = -\frac{\pi}{6}

\frac{11\pi}{6}

So,

z = 2 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right)

Now,

z^{\frac{3}{7}} = (\sqrt{3} - i)^{\frac{3}{7}}

First,

z^3 = (\sqrt{3} - i)^3

= 2^3 \left( \cos\left(\frac{-3\pi}{6}\right) + i \sin\left(\frac{-3\pi}{6}\right) \right)

= 8 \left( \cos\frac{\pi}{2} - i \sin\frac{\pi}{2} \right) \ldots \text{(i)}

Now,

z^{\frac{3}{7}} = (\sqrt{3} - i)^{\frac{3}{7}}

= (8)^{\frac{1}{7}} \left( \cos\left( \frac{-\frac{\pi}{2} + 2k\pi}{7} \right) + i \sin\left( \frac{-\frac{\pi}{2} + 2k\pi}{7} \right) \right) \ldots \text{(ii)}

for

k = 0, 1, \ldots, 6

So,

z^3 = 8(0 - i) = -8i

[Using Eq. (i)]

Now, we know that the product of

\text{nth}

root of a complex number

A

is given by

(-1)^{n-1} A

, if

A

is a real number.

Since,

n = 7

is odd, the product of the 7 roots of

A

A

itself.

\therefore

The product of the 7th roots of

A

A

itself.

\therefore \; (\sqrt{3} - i)^{\frac{3}{7}} = z^3

So, product

= -8i

Go to Question:

Prev Question

Next Question