Math Problem Statement

Find the value of (-1 + i√3)^{10} using De Moivre's Theorem.

Solution

To solve (1+i3)10\left( -1 + i \sqrt{3} \right)^{10} using De Moivre's Theorem, we first need to express the complex number in polar form.

Step 1: Convert to Polar Form

A complex number z=x+iyz = x + iy can be written in polar form as:

z=r(cosθ+isinθ)z = r \left( \cos \theta + i \sin \theta \right)

Where:

  • r=x2+y2r = \sqrt{x^2 + y^2} is the modulus of the complex number,
  • θ=arg(z)=tan1(yx)\theta = \arg(z) = \tan^{-1} \left( \frac{y}{x} \right) is the argument of the complex number.

For z=1+i3z = -1 + i\sqrt{3}:

  • x=1x = -1
  • y=3y = \sqrt{3}
  1. Find the modulus rr: r=(1)2+(3)2=1+3=4=2r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2

  2. Find the argument θ\theta: θ=tan1(31)=tan1(3)\theta = \tan^{-1} \left( \frac{\sqrt{3}}{-1} \right) = \tan^{-1}(-\sqrt{3})

Since tan1(3)\tan^{-1}(-\sqrt{3}) corresponds to an angle of 120120^\circ or 2π3\frac{2\pi}{3} radians in the second quadrant, we have: θ=2π3\theta = \frac{2\pi}{3}

Thus, the polar form of 1+i3-1 + i\sqrt{3} is: z=2(cos2π3+isin2π3)z = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)

Step 2: Apply De Moivre’s Theorem

De Moivre's Theorem states: (r(cosθ+isinθ))n=rn(cos(nθ)+isin(nθ))\left( r \left( \cos \theta + i \sin \theta \right) \right)^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right)

For n=10n = 10, we calculate:

  1. Modulus raised to the power 10: r10=210=1024r^{10} = 2^{10} = 1024

  2. Argument multiplied by 10: nθ=10×2π3=20π3n\theta = 10 \times \frac{2\pi}{3} = \frac{20\pi}{3}

To simplify 20π3\frac{20\pi}{3}, subtract 6π6\pi (equivalent to two full circles) to get: 20π36π=2π3\frac{20\pi}{3} - 6\pi = \frac{2\pi}{3}

Thus, the angle simplifies back to 2π3\frac{2\pi}{3}.

Step 3: Final Expression

Using these values, we get: (1+i3)10=1024(cos2π3+isin2π3)\left( -1 + i\sqrt{3} \right)^{10} = 1024 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)

From trigonometry, we know:

  • cos2π3=12\cos \frac{2\pi}{3} = -\frac{1}{2}
  • sin2π3=32\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}

Therefore: (1+i3)10=1024(12+i32)\left( -1 + i\sqrt{3} \right)^{10} = 1024 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)

Finally: (1+i3)10=512+512i3\left( -1 + i\sqrt{3} \right)^{10} = -512 + 512i\sqrt{3}

Final Answer:

512+512i3\boxed{-512 + 512i\sqrt{3}}

Would you like further clarifications or more details?

Here are five related questions to expand your understanding:

  1. How can we generalize De Moivre's Theorem for any complex number?
  2. What is the geometric significance of multiplying a complex number by itself multiple times?
  3. How would you convert from polar to rectangular form in complex numbers?
  4. How does De Moivre's Theorem relate to Euler's formula?
  5. What happens to the modulus and argument when raising a complex number to a power?

Tip: Always ensure the argument θ\theta lies within the correct quadrant when working with complex numbers in polar form.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Form
De Moivre's Theorem
Trigonometry

Formulas

r = √(x^2 + y^2) (modulus)
θ = tan^{-1}(y/x) (argument)
(r(cosθ + isinθ))^n = r^n(cos(nθ) + isin(nθ)) (De Moivre's Theorem)

Theorems

De Moivre's Theorem

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)