Complex numbers in polar form Knowledge-Based Study

Study notes

1. What is Polar Form?

   A complex number is traditionally written in rectangular (Cartesian) form as:

   $$z = a + bi$$

   where $a$ represents the real part and $b$ represents the imaginary part. In the complex plane, this corresponds to the coordinate point $(a, b)$.

   Alternatively, the position of a complex number can be described using its distance from the origin and the counterclockwise angle it makes with the positive real axis. This representation is known as the polar form (or trigonometric form):

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

   This is often abbreviated as $z = r\operatorname{cis}\theta$.

   Key Components:

   • Modulus ($r$): The distance from the origin to the point $z$. Because distance is strictly non-negative, $r \ge 0$. It is given by:$$r = |z| = \sqrt{a^2 + b^2}$$
   • Argument ($\theta$): The directed angle measured from the positive real axis to the line segment connecting the origin to $z$. The principal argument, denoted $\operatorname{Arg}(z)$, typically lies in the interval $(-\pi, \pi]$ or $[0, 2\pi)$.

   1. From Rectangular to Polar Form

   To convert $z = a + bi$ into polar form $r(\cos\theta + i\sin\theta)$:

   1. Calculate the modulus: $r = \sqrt{a^2 + b^2}$.
   2. Determine the reference angle using $\tan\alpha = \left|\frac{b}{a}\right|$.
   3. Identify the quadrant in which $(a, b)$ lies to find the correct argument $\theta$:
      • Quadrant I: $\theta = \alpha$
      • Quadrant II: $\theta = \pi - \alpha$
      • Quadrant III: $\theta = \pi + \alpha$ (or $-\pi + \alpha$)
      • Quadrant IV: $\theta = 2\pi - \alpha$ (or $-\alpha$)

   Example:

   Convert $z = -1 + i\sqrt{3}$ to polar form with $\theta \in [0, 2\pi)$:

   $$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$

   Since $a < 0$ and $b > 0$, the point lies in Quadrant II. The reference angle is $\arctan(\sqrt{3}) = \frac{\pi}{3}$. Therefore:

   $$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$$$z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$$

   2. From Polar to Rectangular Form

   To convert $z = r(\cos\theta + i\sin\theta)$ back to rectangular form, evaluate the trigonometric functions directly:

   $$a = r\cos\theta, \quad b = r\sin\theta$$

   Operations in Polar Form

   One of the primary advantages of polar form is that multiplication and division become remarkably straightforward geometric operations involving the moduli and arguments.

   Let two complex numbers be given in polar form:

   $$z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$$$$z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$$

   1. Multiplication

   To multiply two complex numbers, multiply their moduli and add their arguments:

   $$z_1 z_2 = r_1 r_2 \left[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right]$$

   Geometric interpretation: Multiplying by a complex number scales the distance from the origin by $r_2$ and rotates the point counterclockwise by angle $\theta_2$.

   2. Division

   To divide two complex numbers (where $z_2 e 0$), divide their moduli and subtract their arguments:

   $$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right]$$

   Worked Example:

   Let $z_1 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$ and $z_2 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$.

   $$z_1 z_2 = (6 \cdot 2)\left[\cos\left(\frac{\pi}{2} + \frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{2} + \frac{\pi}{6}\right)\right] = 12\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$$$$\frac{z_1}{z_2} = \frac{6}{2}\left[\cos\left(\frac{\pi}{2} - \frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{2} - \frac{\pi}{6}\right)\right] = 3\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$$

   De Moivre's Theorem

   Extending the multiplication rule to repeated multiplication yields De Moivre's Theorem, which provides an efficient method for raising complex numbers to integer powers.

   For any complex number $z = r(\cos\theta + i\sin\theta)$ and any integer $n$:

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

   Example: Computing Powers

   Compute $(1 + i)^8$ using polar form:

   1. Modulus: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$
   2. Argument: $\theta = \frac{\pi}{4}$ (Quadrant I)
   3. Apply De Moivre's Theorem:$$(1 + i)^8 = (\sqrt{2})^8 \left[\cos\left(8 \cdot \frac{\pi}{4}\right) + i\sin\left(8 \cdot \frac{\pi}{4}\right)\right]$$$$= 16 (\cos 2\pi + i\sin 2\pi) = 16(1 + 0i) = 16$$

   Roots of Complex Numbers

   To find the $n$ distinct $n$-th roots of $z = r(\cos\theta + i\sin\theta)$, use the formula:

   $$w_k = r^{1/n}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right]$$

   for $k = 0, 1, 2, \dots, n-1$. In the complex plane, these $n$ roots are evenly spaced along a circle of radius $r^{1/n}$, forming the vertices of a regular $n$-sided polygon centered at the origin.