State De Moivre's theorem. Using De Moivre's theorem, find the cube roots of unity. If w = 1+ 3i2 be a complex cube root of unity, show that w^2 = 1 3i2 .

De Moivre’s Theorem and Cube Roots of Unity

1. Statement of De Moivre’s Theorem

De Moivre’s Theorem states that for any integer $n$ and a complex number $z=r(\cos \theta +i\sin \theta )$, the following holds:

$$z^n=r^n(\cos (n\theta )+i\sin (n\theta ))$$

This theorem is particularly useful for raising complex numbers to powers and finding roots of complex numbers.

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2. Finding the Cube Roots of Unity Using De Moivre’s Theorem

The cube roots of unity are the solutions to the equation:

$$x^3=1$$

In polar form, $1$ can be written as:

$$1=1(\cos 0+i\sin 0)$$

Using De Moivre’s Theorem, the cube roots of unity are given by:

$$x_k=1^{1/3}\left[\cos \left(\frac{0+2k\pi }{3}\right)+i\sin \left(\frac{0+2k\pi }{3}\right)\right],k=0,1,2$$

Thus, the three cube roots of unity are:

1. First root ($k=0$): $$x_0=\cos 0+i\sin 0=1$$

2. Second root ($k=1$): $$x_1=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$$

3. Third root ($k=2$): $$x_2=\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$

So, the cube roots of unity are: $$\boxed{1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}}$$

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3. Verification of $w^2$ for a Given Cube Root of Unity

Given: $$w=\frac{-1+\sqrt{3}i}{2}$$

We need to show that: $$w^2=\frac{-1-\sqrt{3}i}{2}$$

Step 1: Express $w$ in Polar Form

First, write $w$ in the form $r(\cos \theta +i\sin \theta )$.

• Magnitude ($r$): $$r=\sqrt{{\left(-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1$$

• Argument ($\theta$): Since $w$ lies in the second quadrant (negative real part, positive imaginary part), $$\theta =\pi -{\tan }^{-1}\left(\left|\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right|\right)=\pi -{\tan }^{-1}(\sqrt{3})=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$$

Thus, $$w=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)$$

Step 2: Compute $w^2$ Using De Moivre’s Theorem

Using De Moivre’s Theorem: $$w^2={\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right]}^2=\cos \left(2\cdot \frac{2\pi }{3}\right)+i\sin \left(2\cdot \frac{2\pi }{3}\right)$$ $$=\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)$$

Now, evaluate the trigonometric functions: $$\cos \left(\frac{4\pi }{3}\right)=-\frac{1}{2},\sin \left(\frac{4\pi }{3}\right)=-\frac{\sqrt{3}}{2}$$

Thus, $$w^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}=\frac{-1-\sqrt{3}i}{2}$$

This matches the given expression, proving the result.

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4. Summary of Cube Roots of Unity

The three cube roots of unity are:

1. $1$ (real root)
2. $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
3. ${\omega }^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

These roots satisfy the property: $$1+\omega +{\omega }^2=0$$