Finding the Square Root of \(-\sqrt{-1}\)

The question asks us to find a square root of the complex number \(-\sqrt{-1}\). First, let's understand what \(-\sqrt{-1}\) represents. We know that the imaginary unit, denoted by \(i\), is defined as \(i = \sqrt{-1}\).

Therefore, \(-\sqrt{-1}\) is simply equal to \(-i\). So, the problem reduces to finding the square root(s) of \(-i\).

A square root of a complex number \(w\) is a complex number \(z\) such that \(z^2 = w\).

We will use two methods to find the square roots of \(-i\):

Method 1: Algebraic Method

Let the square root of \(-i\) be \(z = x + iy\), where \(x\) and \(y\) are real numbers. Squaring \(z\), we get:

\(z^2 = (x + iy)^2 = x^2 + 2xy(i) + (iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + i(2xy)\)

We are looking for \(z^2 = -i\). So, we set the squared form equal to \(-i\):

\((x^2 - y^2) + i(2xy) = 0 - 1i\)

By comparing the real and imaginary parts of this equation, we get a system of two real equations:

1. \(x^2 - y^2 = 0\)
2. \(2xy = -1\)

From equation (1), \(x^2 = y^2\), which means \(y = x\) or \(y = -x\).

• Consider the case \(y = x\). Substitute this into equation (2):

  \(2x(x) = -1\)

  \(2x^2 = -1\)

  \(x^2 = -\frac{1}{2}\)

  Since \(x\) must be a real number, \(x^2\) cannot be negative. Thus, there are no real solutions for \(x\) in this case. This means \(y\) cannot be equal to \(x\).

• Consider the case \(y = -x\). Substitute this into equation (2):

  \(2x(-x) = -1\)

  \(-2x^2 = -1\)

  \(2x^2 = 1\)

  \(x^2 = \frac{1}{2}\)

  This gives two possible real values for \(x\): \(x = \frac{1}{\sqrt{2}}\) or \(x = -\frac{1}{\sqrt{2}}\).

  • If \(x = \frac{1}{\sqrt{2}}\), then \(y = -x = -\frac{1}{\sqrt{2}}\).

    The square root is \(z_1 = x + iy = \frac{1}{\sqrt{2}} + i\left(-\frac{1}{\sqrt{2}}\right) = \frac{1 - i}{\sqrt{2}}\).

  • If \(x = -\frac{1}{\sqrt{2}}\), then \(y = -x = -\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\).

    The square root is \(z_2 = x + iy = -\frac{1}{\sqrt{2}} + i\left(\frac{1}{\sqrt{2}}\right) = \frac{-1 + i}{\sqrt{2}}\).

So the two square roots of \(-i\) are \(\frac{1-i}{\sqrt{2}}\) and \(\frac{-1+i}{\sqrt{2}}\).

Method 2: Polar Form Method

We want to find the square roots of \(-i\). Let's express \(-i\) in polar form, \(r(\cos \theta + i \sin \theta)\).

The modulus \(r\) of \(-i\) is \(|-i| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1\).

The argument \(\theta\) of \(-i\) is the angle such that \(\cos \theta = \frac{0}{1} = 0\) and \(\sin \theta = \frac{-1}{1} = -1\). A suitable angle in the range \((-\pi, \pi]\) is \(\theta = -\frac{\pi}{2}\). Another possible angle is \(\frac{3\pi}{2}\).

So, \(-i = 1\left(\cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)\right)\).

The square roots of a complex number \(r(\cos \theta + i \sin \theta)\) are given by De Moivre's theorem for roots:

\(\sqrt{r}\left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right)\right)\), where \(n\) is the root we are looking for (here \(n=2\) for square roots) and \(k\) takes values \(0, 1, \dots, n-1\).

For the square roots of \(-i\), we have \(r=1\), \(\theta = -\frac{\pi}{2}\), and \(n=2\). The values for \(k\) are \(0\) and \(1\).

• For \(k=0\):

  \(z_1 = \sqrt{1}\left(\cos\left(\frac{-\frac{\pi}{2} + 2(0)\pi}{2}\right) + i \sin\left(\frac{-\frac{\pi}{2} + 2(0)\pi}{2}\right)\right)\)

  \(z_1 = 1\left(\cos\left(\frac{-\frac{\pi}{2}}{2}\right) + i \sin\left(\frac{-\frac{\pi}{2}}{2}\right)\right)\)

  \(z_1 = \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)\)

  \(z_1 = \frac{1}{\sqrt{2}} + i\left(-\frac{1}{\sqrt{2}}\right) = \frac{1 - i}{\sqrt{2}}\)

• For \(k=1\):

  \(z_2 = \sqrt{1}\left(\cos\left(\frac{-\frac{\pi}{2} + 2(1)\pi}{2}\right) + i \sin\left(\frac{-\frac{\pi}{2} + 2(1)\pi}{2}\right)\right)\)

  \(z_2 = \cos\left(\frac{-\frac{\pi}{2} + \frac{4\pi}{2}}{2}\right) + i \sin\left(\frac{-\frac{\pi}{2} + \frac{4\pi}{2}}{2}\right)\)

  \(z_2 = \cos\left(\frac{\frac{3\pi}{2}}{2}\right) + i \sin\left(\frac{\frac{3\pi}{2}}{2}\right)\)

  \(z_2 = \cos\left(\frac{3\pi}{4}\right) + i \sin\left(\frac{3\pi}{4}\right)\)

  \(z_2 = -\frac{1}{\sqrt{2}} + i\left(\frac{1}{\sqrt{2}}\right) = \frac{-1 + i}{\sqrt{2}}\)

Both methods give the same two square roots of \(-i\): \(\frac{1-i}{\sqrt{2}}\) and \(\frac{-1+i}{\sqrt{2}}\).

Checking the Options

Now let's compare our results with the given options to find which one is a square root of \(-\sqrt{-1}\) (which is \(-i\)).

Based on our calculations, the option that is a square root of \(-\sqrt{-1}\) is \(\frac{1-i}{\sqrt{2}}\).

Revision Table: Key Concepts for Complex Number Roots

Additional Information: Understanding Square Roots of Complex Numbers

Every non-zero complex number has exactly two distinct square roots. These two roots are always negatives of each other.

For example, the square roots of \(-i\) are \(\frac{1-i}{\sqrt{2}}\) and \(\frac{-1+i}{\sqrt{2}}\). Notice that \(\frac{-1+i}{\sqrt{2}} = -1 \times \left(\frac{1-i}{\sqrt{2}}\right)\).

Graphically, if a complex number \(z\) has modulus \(r\) and argument \(\theta\), its square roots have modulus \(\sqrt{r}\). Their arguments are \(\frac{\theta}{2}\) and \(\frac{\theta}{2} + \pi\). This means the two square roots are located diametrically opposite to each other on a circle of radius \(\sqrt{r}\) centered at the origin in the complex plane.

For \(-i\), the modulus is 1, and a principal argument is \(-\frac{\pi}{2}\). The modulus of the square roots is \(\sqrt{1}=1\). The arguments are \(\frac{-\pi/2}{2} = -\frac{\pi}{4}\) and \(\frac{-\pi/2}{2} + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}\). The complex numbers with modulus 1 and arguments \(-\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) are \(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\) and \(\cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\), which matches our results.