Question

Number of real values of $(-1-\sqrt{3}i)^{3/4}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

To find the $n$-th roots of a complex number, always use its general polar form $z=r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi))$ and then apply De Moivre's theorem for fractional exponents. The distinct roots are found by using $k=0, 1, 2, ..., n-1$. A root is real if its imaginary part ($\sin$ term) is zero.

Answer 1

The Correct Option is C

Let $z = -1 - \sqrt{3}i$. We want to find the number of real values of $z^{3/4}$.
First, convert $z$ to its polar form, $z = r(\cos\theta + i\sin\theta)$.
Modulus $r = |z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.
Argument $\theta = \arctan(\frac{-\sqrt{3}}{-1}) = \arctan(\sqrt{3})$. Since both x and y are negative, the angle is in the third quadrant. So, $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
The general polar form is $z = 2(\cos(\frac{4\pi}{3} + 2k\pi) + i\sin(\frac{4\pi}{3} + 2k\pi))$ for $k \in \mathbb{Z}$.
Now, we find $z^{3/4}$ using De Moivre's theorem:
$z^{3/4} = (2)^{3/4} \left( \cos\left(\frac{3}{4}(\frac{4\pi}{3} + 2k\pi)\right) + i\sin\left(\frac{3}{4}(\frac{4\pi}{3} + 2k\pi)\right) \right)$.
$z^{3/4} = \sqrt[4]{8} \left( \cos\left(\pi + \frac{3k\pi}{2}\right) + i\sin\left(\pi + \frac{3k\pi}{2}\right) \right)$.
We will get four distinct roots for $k=0, 1, 2, 3$. For a value to be real, its imaginary part must be zero.
We need $\sin(\pi + \frac{3k\pi}{2}) = 0$.
The sine function is zero at integer multiples of $\pi$. So, we need $\pi + \frac{3k\pi}{2} = m\pi$ for some integer $m$.
$1 + \frac{3k}{2} = m \implies 2 + 3k = 2m$. This means $3k$ must be an even number, which implies $k$ must be even.
Let's check the values of $k=0, 1, 2, 3$:
For $k=0$ (which is even): The angle is $\pi$. $\sin(\pi) = 0$. The value is $\sqrt[4]{8}(\cos(\pi)) = -\sqrt[4]{8}$. This is a real value.
For $k=1$: The angle is $\pi + \frac{3\pi}{2} = \frac{5\pi}{2}$. $\sin(\frac{5\pi}{2})=1$. Not real.
For $k=2$ (which is even): The angle is $\pi + \frac{3(2)\pi}{2} = \pi + 3\pi = 4\pi$. $\sin(4\pi) = 0$. The value is $\sqrt[4]{8}(\cos(4\pi)) = \sqrt[4]{8}$. This is a real value.
For $k=3$: The angle is $\pi + \frac{3(3)\pi}{2} = \frac{11\pi}{2}$. $\sin(\frac{11\pi}{2})=-1$. Not real.
So, there are two real values corresponding to $k=0$ and $k=2$.