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To solve problems involving complex numbers, it's often useful to express them in polar form. This is because polar form simplifies many operations, such as finding roots.

A complex number in standard form is expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. Its polar form is written as \( r(\cos \theta + i\sin \theta) \), or more compactly, \( r e^{i\theta} \).

Here, \( r \) represents the modulus (or absolute value) of the complex number, and \( \theta \) is the argument (or angle).

To convert a complex number to polar form:

• Calculate the modulus: \( r = \sqrt{a^2 + b^2} \).
• Find the argument: \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

In the exercise, for \( n + ni\sqrt{3} \), the modulus is \( 2n \) and the argument is \( \frac{\pi}{3} \). This gives the polar form: \( 2n( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} ) \).

De Moivre's Theorem is a powerful tool when dealing with complex numbers, especially when you need to find powers or roots.

The theorem states: For any complex number in polar form \( r(\cos \theta + i \sin \theta) \) and any integer \( n \), \( [r(\cos \theta + i \sin \theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta)) \).

This theorem also helps in finding the nth roots of a complex number. For identifying square roots, when \( n = 2 \), this gives us:

• \( \sqrt{r}(\cos \frac{\theta + 2k\pi}{2} + i\sin \frac{\theta + 2k\pi}{2}) \), for \( k = 0, 1 \).

Using De Moivre's Theorem in the context of the exercise, \( r = 2n \). Therefore, \( \sqrt{r} = \sqrt{2n} \).
This formulation helps compute all necessary roots easily.

Finding square roots of complex numbers can be tricky, but using polar form along with De Moivre's Theorem makes the process straightforward.

Once the complex number is in polar form, use De Moivre's Theorem to calculate the square roots. This involves determining two distinct roots by varying \( k \) in the theorem from 0 to 1.

For polar representation \( r(\cos \theta + i\sin \theta) \):

• First root for \( k = 0 \): \( \sqrt{r}(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}) \)
• Second root for \( k = 1 \): \( \sqrt{r}(\cos \frac{\theta + \pi}{2} + i \sin \frac{\theta + \pi}{2}) \)

In the exercise:

• For \( k = 0 \), the root is \( \sqrt{2n}(\frac{\sqrt{3}}{2} + i\frac{1}{2}) \).
• For \( k = 1 \), the root is \( \sqrt{2n}(-\frac{\sqrt{3}}{2} - i\frac{1}{2}) \).

These steps provide the complete solution to finding the square roots of \( n + ni\sqrt{3} \).