Complex Numbers and De Moivre's Theorem

Forms of Complex Numbers

• Polar Form: $r \cdot cis(\theta)$, where $cis(\theta) = cos(\theta) + i sin(\theta)$

  • $r$ is the modulus of $Z$ and represents the distance to the origin.

• Exponential Form: $r \cdot e^{i\theta}$ where $r$ is the modulus and $\theta$ is the angle.

• Moving between polar and exponential forms involves recognizing the direct correspondence between components.

Moving Between Forms

• Standard Form to Coordinate Form: Easy; identify the real and imaginary parts.

• Polar to Standard Form: Multiply out the polar form: $r \cdot cos(\theta)$ is the real part, and $r \cdot sin(\theta)$ is the imaginary part.

• Angles are typically special ratios (e.g., $\frac{\pi}{4}$, $\frac{\pi}{6}$, $\frac{\pi}{3}$) derived from special triangles or axes.

• Standard to Polar Form:

  • $r = |Z| = \sqrt{a^2 + b^2}$

  • $\theta$ is found using $tan(\theta) = \frac{b}{a}$; consider the quadrant based on the signs of $a$ and $b$.

  • $\theta$ must start from the positive real axis, with counter-clockwise being positive and clockwise negative.

Argument of Complex Number

• The principal argument, $\theta$, lies in the interval $(-\pi, \pi]$.

• Adjust $\theta$ by adding or subtracting revolutions ($2k\pi$) to fit within the interval.

• For any trigonometric function, $trig(\theta) = trig(\theta + 2k\pi)$, where $k$ is an integer.

Products and Division in Polar Form

• To multiply complex numbers, multiply the moduli and add the angles: $Z<em>1 \cdot Z</em>2$, then $r<em>1 \cdot r</em>2$ and $\theta<em>1 + \theta</em>2$.

• To divide complex numbers, divide the moduli and subtract the angles: $\frac{Z<em>1}{Z</em>2}$, then $\frac{r<em>1}{r</em>2}$ and $\theta<em>1 - \theta</em>2$.

Power Rule

• $Z^n = r^n \cdot cis(n\theta)$, where $n$ is an integer.

• This can be further understood using exponential form: $(r<em>1 e^{i\theta</em>1}) \cdot (r<em>2 e^{i\theta</em>2}) = r<em>1 r</em>2 e^{i(\theta<em>1 + \theta</em>2)}$.

De Moivre's Theorem

• If $Z = r \cdot cis(\theta)$, then $Z^n = r^n \cdot cis(n\theta)$, where $n$ is an integer.

  • Case 1: $n = 0$. Then $Z^0 = 1$ and $r^0 \cdot cis(0) = 1$.

  • Case 2: n > 0. Use mathematical induction.

    • Base case: $n = 1$. $Z^1 = r \cdot cis(\theta)$.

    • Assume true for $n = k$. Then $Z^k = r^k \cdot cis(k\theta)$.

    • Prove for $n = k + 1$. $Z^{k+1} = Z^k \cdot Z = r^k \cdot cis(k\theta) \cdot r \cdot cis(\theta) = r^{k+1} \cdot cis((k+1)\theta)$.

  • Case 3: n < 0. Let $n = -m$, where m > 0.

    • $Z^n = Z^{-m} = \frac{1}{Z^m} = \frac{1}{r^m \cdot cis(m\theta)}$. Rationalize to simplify.

Proof of Multiplication of Complex Numbers

• Given $Z<em>1 = r</em>1(cos(\theta<em>1) + i sin(\theta</em>1))$ and $Z<em>2 = r</em>2(cos(\theta<em>2) + i sin(\theta</em>2))$, then
  $Z<em>1 \cdot Z</em>2 = r<em>1 r</em>2 [cos(\theta<em>1)cos(\theta</em>2) - sin(\theta<em>1)sin(\theta</em>2) + i(sin(\theta<em>1)cos(\theta</em>2) + cos(\theta<em>1)sin(\theta</em>2))] = r<em>1 r</em>2 [cos(\theta<em>1 + \theta</em>2) + i sin(\theta<em>1 + \theta</em>2)]$.

Multiplication Example

• Convert complex numbers to polar form, such as $Z<em>1 = 5\sqrt{10} \cdot cis(\frac{3\pi}{4})$ and $Z</em>2 = 2 \cdot cis(-\frac{\pi}{6})$.

• Multiply moduli and add arguments, adjusting the angle to fit the principal argument interval.

Division Example

• To divide, divide the moduli and subtract the angles, ensuring the result is within the principal argument range.

Power Example

• Raise $Z<em>1$ to the power of four: $Z</em>1^4 = (5\sqrt{2})^4 \cdot cis(3\pi)$. Simplify and adjust the angle by subtracting $2\pi$ to get it within the interval.
  Converting to Standard Form
  To convert from polar to standard form, compute the real and imaginary parts using cosine and sine, respectively.

Extracting Roots of Complex Numbers

• Solving $x^2 = 4$ involves factorizing and finding roots.

• Equations of degree $n$ typically have $n$ roots.

• The roots of complex numbers can be found using the formula in Theorem 59.

Theorem 59

• The equation $Z^n = r \cdot cis(\theta)$ has $n$ different complex roots of the form

  $W_k = r^{\frac{1}{n}} \cdot cis(\frac{\theta + 2k\pi}{n})$ for $k = 0, 1, 2, …, n-1$.

Root Extraction Example

• To solve $Z^3 = 1 + i$, first convert $1 + i$ to polar form.

  • $r = \sqrt{1^2 + 1^2} = \sqrt{2}$. $\theta = tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}$.

• Apply the formula to find the cube roots,$W_k$ , for $k = 0, 1, 2$.

• Simplify the roots and ensure the arguments are principal.

Detailed Steps for Root Extraction

• Express $Z = 1 + i$ in polar form: $\sqrt{2} \cdot cis(\frac{\pi}{4})$.

• The roots are $W_k = (\sqrt{2})^{\frac{1}{3}} \cdot cis(\frac{\frac{\pi}{4} + 2k\pi}{3})$.

Calculations for Each Root

• For $k = 0$: $W_0 = 2^{\frac{1}{6}} \cdot cis(\frac{\pi}{12})$.

• For $k = 1$: $W_1 = 2^{\frac{1}{6}} \cdot cis(\frac{\frac{\pi}{4} + 2\pi}{3}) = 2^{\frac{1}{6}} \cdot cis(\frac{3\pi}{4})$.

• For $k = 2$: $W_2 = 2^{\frac{1}{6}} \cdot cis(\frac{\frac{\pi}{4} + 4\pi}{3})$. This simplifies to $2^{\frac{1}{6}} \cdot cis(-\frac{7\pi}{12})$ after adjusting by subtracting $2\pi$.

Visual Representation of Roots

• The roots are equally spaced around a circle in the complex plane.

• The angles between roots are equal, demonstrating the properties of root extraction.

• If complex number is in the form of $a + bi$, it should not be written as $\frac{a - bi}{c}$; instead write it as ($\frac{a}{c} - \frac{bi}{c}$).

Combining Theorems

Solve for $Z^3 = (1+i)^2$: first apply De Moivre's theorem to $(1+i)^2$, then extract the roots.