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To simplify [tex]\((\sqrt{3} + i)^3\)[/tex] using De Moivre's theorem, we'll follow a step-by-step approach. Here’s how:
1. Express the Complex Number in Polar Form:
A complex number [tex]\( z = \sqrt{3} + i \)[/tex] can be rewritten in polar form [tex]\( z = r (\cos \theta + i \sin \theta) \)[/tex], where [tex]\( r \)[/tex] is the modulus and [tex]\( \theta \)[/tex] is the argument (or angle).
2. Find the Modulus [tex]\( r \)[/tex]:
The modulus [tex]\( r \)[/tex] is given by:
[tex]\[ r = |z| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \][/tex]
3. Find the Argument [tex]\( \theta \)[/tex]:
The argument [tex]\( \theta \)[/tex] is the angle made with the positive real axis and is given by:
[tex]\[ \theta = \arctan \left( \frac{\text{Imaginary part}}{\text{Real part}} \right) = \arctan \left( \frac{1}{\sqrt{3}} \right). \][/tex]
Since [tex]\(\arctan \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \)[/tex], we have:
[tex]\[ \theta = \frac{\pi}{6}. \][/tex]
4. Write the Complex Number in Polar Form:
Now we can write the complex number [tex]\(\sqrt{3} + i\)[/tex] as:
[tex]\[ \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right). \][/tex]
5. Use De Moivre's Theorem:
De Moivre's theorem states that for a complex number [tex]\( z = r (\cos \theta + i \sin \theta) \)[/tex] and an integer [tex]\( n \)[/tex]:
[tex]\[ z^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Here, [tex]\( n = 3 \)[/tex], so:
[tex]\[ (\sqrt{3} + i)^3 = \left[ 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right]^3. \][/tex]
First calculate:
[tex]\[ r^3 = 2^3 = 8. \][/tex]
Now, calculate the new argument:
[tex]\[ n \theta = 3 \theta = 3 \cdot \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}. \][/tex]
6. Evaluate the Trigonometric Functions:
[tex]\[ \cos \frac{\pi}{2} = 0 \quad \text{and} \quad \sin \frac{\pi}{2} = 1. \][/tex]
7. Write the Result in Rectangular Form:
Substitute the values back in:
[tex]\[ (\sqrt{3} + i)^3 = 8 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = 8 (0 + i \cdot 1) = 8i. \][/tex]
Therefore, the simplified form of [tex]\((\sqrt{3} + i)^3\)[/tex] is:
[tex]\[ (\sqrt{3} + i)^3 = 0 + 8i. \][/tex]
Putting it into the form [tex]\(\square + \square i\)[/tex]:
[tex]\[ (\sqrt{3} + i)^3 = 0 + 8i. \][/tex]
1. Express the Complex Number in Polar Form:
A complex number [tex]\( z = \sqrt{3} + i \)[/tex] can be rewritten in polar form [tex]\( z = r (\cos \theta + i \sin \theta) \)[/tex], where [tex]\( r \)[/tex] is the modulus and [tex]\( \theta \)[/tex] is the argument (or angle).
2. Find the Modulus [tex]\( r \)[/tex]:
The modulus [tex]\( r \)[/tex] is given by:
[tex]\[ r = |z| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \][/tex]
3. Find the Argument [tex]\( \theta \)[/tex]:
The argument [tex]\( \theta \)[/tex] is the angle made with the positive real axis and is given by:
[tex]\[ \theta = \arctan \left( \frac{\text{Imaginary part}}{\text{Real part}} \right) = \arctan \left( \frac{1}{\sqrt{3}} \right). \][/tex]
Since [tex]\(\arctan \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \)[/tex], we have:
[tex]\[ \theta = \frac{\pi}{6}. \][/tex]
4. Write the Complex Number in Polar Form:
Now we can write the complex number [tex]\(\sqrt{3} + i\)[/tex] as:
[tex]\[ \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right). \][/tex]
5. Use De Moivre's Theorem:
De Moivre's theorem states that for a complex number [tex]\( z = r (\cos \theta + i \sin \theta) \)[/tex] and an integer [tex]\( n \)[/tex]:
[tex]\[ z^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Here, [tex]\( n = 3 \)[/tex], so:
[tex]\[ (\sqrt{3} + i)^3 = \left[ 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right]^3. \][/tex]
First calculate:
[tex]\[ r^3 = 2^3 = 8. \][/tex]
Now, calculate the new argument:
[tex]\[ n \theta = 3 \theta = 3 \cdot \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}. \][/tex]
6. Evaluate the Trigonometric Functions:
[tex]\[ \cos \frac{\pi}{2} = 0 \quad \text{and} \quad \sin \frac{\pi}{2} = 1. \][/tex]
7. Write the Result in Rectangular Form:
Substitute the values back in:
[tex]\[ (\sqrt{3} + i)^3 = 8 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = 8 (0 + i \cdot 1) = 8i. \][/tex]
Therefore, the simplified form of [tex]\((\sqrt{3} + i)^3\)[/tex] is:
[tex]\[ (\sqrt{3} + i)^3 = 0 + 8i. \][/tex]
Putting it into the form [tex]\(\square + \square i\)[/tex]:
[tex]\[ (\sqrt{3} + i)^3 = 0 + 8i. \][/tex]
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