Complex number - Polar Form and Power Operations

Polar Representation and Functions Introduction So far, we've worked with complex numbers in rectangular form: $z = a + bi$. However, there's another powerful way to represent complex numbers using their magnitude and angle—the polar form. This representation transforms complex multiplication and division into simple operations on magnitudes and angles. More importantly, it provides elegant formulas for computing powers and roots of complex numbers. In this section, we'll explore polar coordinates and their applications. Polar Coordinates: A New Perspective Every non-zero complex number $z = a + bi$ can be uniquely represented by two values: The modulus (or absolute value): $r = |z| = \sqrt{a^2 + b^2}$, which measures the distance from the origin The argument: $\theta = \arg(z)$, which measures the angle from the positive real axis (measured counterclockwise) With these two values, we can express any complex number as: $$z = r(\cos\theta + i\sin\theta)$$ This is called the polar form of the complex number. Important note about the argument: The argument is defined modulo $2\pi$, meaning that $\theta$ and $\theta + 2\pi$ represent the same angle. To avoid confusion, we use the principal value of the argument, which is the unique value satisfying $-\pi < \theta \leq \pi$. This ensures each complex number has exactly one principal argument. Euler's Formula: A Bridge Between Forms The most elegant relationship in complex analysis is Euler's formula: $$e^{i\theta} = \cos\theta + i\sin\theta$$ This remarkable identity connects exponential functions to trigonometric functions through the imaginary unit. Using this formula, we can write complex numbers in exponential form: $$z = re^{i\theta}$$ where $r = |z|$ and $\theta = \arg(z)$. Both forms—$r(\cos\theta + i\sin\theta)$ and $re^{i\theta}$—represent exactly the same complex number. The exponential form is often more convenient for calculations because it turns trigonometric identities into laws of exponents. Example: Let's express $z = 1 + i$ in polar form. The modulus is $r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$. The argument is $\theta = \arctan(1/1) = \pi/4$. Therefore: $$z = 1 + i = \sqrt{2}e^{i\pi/4} = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$$ Operations in Polar Form One of the major advantages of polar form is that multiplication and division become remarkably simple. Multiplication When we multiply $z1 = r1 e^{i\theta1}$ and $z2 = r2 e^{i\theta2}$, we get: $$z1 z2 = (r1 r2) e^{i(\theta1 + \theta2)}$$ In other words: To multiply two complex numbers in polar form, multiply their moduli and add their arguments. Example: Let $z1 = 2e^{i\pi/6}$ and $z2 = 3e^{i\pi/3}$. Then: $$z1 z2 = 6e^{i(\pi/6 + \pi/3)} = 6e^{i\pi/2} = 6i$$ Division Similarly, dividing $z1$ by $z2$ (where $z2 \neq 0$) yields: $$\frac{z1}{z2} = \frac{r1}{r2} e^{i(\theta1 - \theta2)}$$ In other words: To divide two complex numbers in polar form, divide their moduli and subtract their arguments. Example: Let $z1 = 6e^{i\pi/2}$ and $z2 = 2e^{i\pi/6}$. Then: $$\frac{z1}{z2} = \frac{6}{2}e^{i(\pi/2 - \pi/6)} = 3e^{i\pi/3}$$ Compare this to multiplying or dividing in rectangular form—it's far messier! This is one reason why polar form is so useful. Computing Arguments To work with polar form, you need to be able to compute the argument of a complex number $z = a + bi$. The argument can be found using the two-argument arctangent function, denoted $\operatorname{atan2}(b, a)$: $$\arg(z) = \operatorname{atan2}(b, a)$$ This function is preferable to simply computing $\arctan(b/a)$ because it correctly accounts for which quadrant the complex number lies in, automatically giving you the principal value in the range $(-\pi, \pi]$. Quick guide without a calculator: If $a > 0$ and $b \geq 0$ (first quadrant): $\theta = \arctan(b/a)$ If $a \leq 0$ and $b > 0$ (second quadrant): $\theta = \pi - \arctan(b/|a|)$ If $a < 0$ and $b \leq 0$ (third quadrant): $\theta = -\pi + \arctan(|b|/|a|)$ If $a \geq 0$ and $b < 0$ (fourth quadrant): $\theta = -\arctan(|b|/a)$ De Moivre's Formula: Powers of Complex Numbers De Moivre's formula is a powerful tool for computing powers of complex numbers: $$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$ for any integer $n$. In exponential form: $$z^n = r^n e^{in\theta}$$ Why is this useful? Computing $(a + bi)^n$ directly by repeated multiplication becomes increasingly tedious as $n$ grows. De Moivre's formula reduces this to simple arithmetic: raise the modulus to the $n$-th power and multiply the argument by $n$. Example: Let's compute $(1 + i)^3$ using De Moivre's formula. First, convert to polar form: $1 + i = \sqrt{2}e^{i\pi/4}$. Then: $$(1 + i)^3 = (\sqrt{2})^3 e^{i(3\pi/4)} = 2\sqrt{2}e^{i3\pi/4} = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$$ $$= 2\sqrt{2}\left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) = -2 + 2i$$ Notice how much cleaner this is than expanding $(1 + i)^3$ by hand! Powers of the Imaginary Unit A simple but important application of De Moivre's formula is the pattern of powers of $i$: $$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \ldots$$ The powers repeat with period 4. So to compute $i^n$ for any integer $n$, find the remainder when $n$ is divided by 4: If $n \equiv 0 \pmod{4}$: $i^n = 1$ If $n \equiv 1 \pmod{4}$: $i^n = i$ If $n \equiv 2 \pmod{4}$: $i^n = -1$ If $n \equiv 3 \pmod{4}$: $i^n = -i$ Finding n-th Roots One of the most important applications of De Moivre's formula is finding $n$-th roots of complex numbers. Given a non-zero complex number $z = re^{i\theta}$, the $n$-th roots are: $$\sqrt[n]{z} = r^{1/n}\left(\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n}\right)$$ for $k = 0, 1, 2, \ldots, n-1$. This formula tells us something remarkable: every non-zero complex number has exactly $n$ distinct $n$-th roots. Why do we have $n$ roots? The key is that adding multiples of $2\pi$ to the argument doesn't change the original number (since arguments are defined modulo $2\pi$), but it produces different $n$-th roots. By letting $k$ range from $0$ to $n-1$, we generate all $n$ distinct roots. Example: Find the cube roots of $8$ (which is $8e^{i \cdot 0}$ in polar form). Here, $r = 8$, $\theta = 0$, and $n = 3$. The three cube roots are: $$\sqrt[3]{8} = 2\left(\cos\frac{0 + 2k\pi}{3} + i\sin\frac{0 + 2k\pi}{3}\right) \quad \text{for } k = 0, 1, 2$$ $k = 0$: $2(\cos 0 + i\sin 0) = 2$ $k = 1$: $2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$ $k = 2$: $2\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}$ Notice that these three roots are evenly spaced around a circle of radius $2$ in the complex plane. This is true in general: the $n$ roots of any complex number are evenly distributed on a circle. Geometric insight: The $n$-th roots of $z = re^{i\theta}$ all have modulus $r^{1/n}$ (they lie on a circle), and their arguments differ by $2\pi/n$ (they're evenly spaced).