2.4 Complex Numbers

Complex Numbers

Complex numbers extend the real number system so that every polynomial equation has a solution. Without them, an equation like $x^2 + 1 = 0$ simply has no answer. A complex number takes the form $a + bi$, where $a$ is the real part, $b$ is the imaginary part, and $i$ is defined as $i = \sqrt{-1}$.

A few things worth knowing upfront:

• Every real number is technically a complex number with $b = 0$. For example, $3$ is really $3 + 0i$.
• A pure imaginary number has $a = 0$, like $0 + 2i$ (usually just written $2i$).
• Two complex numbers are equal only when their real parts match and their imaginary parts match.

Arithmetic with Complex Numbers

Addition and subtraction work just like combining like terms. You handle the real parts and imaginary parts separately.

• $(2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i$
• $(7 - 2i) - (3 + 4i) = (7 - 3) + (-2 - 4)i = 4 - 6i$

Multiplication uses the distributive property (FOIL works here). The key rule is $i^2 = -1$, so whenever you get an $i^2$ term, replace it with $-1$.

• $(2 + 3i)(4 + 5i) = 8 + 10i + 12i + 15i^2 = 8 + 22i + 15(-1) = -7 + 22i$
• $(1 - i)(2 + 3i) = 2 + 3i - 2i - 3i^2 = 2 + i - 3(-1) = 5 + i$

Division requires a special trick. You multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of $a + bi$ is $a - bi$. This eliminates $i$ from the denominator because $(a + bi)(a - bi) = a^2 + b^2$, which is always a real number.

Step-by-step for $\frac{2 + 3i}{4 + 5i}$:

1. Identify the conjugate of the denominator: $4 - 5i$

2. Multiply top and bottom by that conjugate: $\frac{(2 + 3i)(4 - 5i)}{(4 + 5i)(4 - 5i)}$

3. Expand the numerator: $8 - 10i + 12i - 15i^2 = 8 + 2i + 15 = 23 + 2i$

4. Expand the denominator: $16 - 25i^2 = 16 + 25 = 41$

5. Write in standard form: $\frac{23}{41} + \frac{2}{41}i$

Powers of the Imaginary Unit

The powers of $i$ follow a cyclic pattern that repeats every four:

After $i^4$, the cycle starts over: $i^5 = i$, $i^6 = -1$, and so on.

To simplify any power of $i$, divide the exponent by 4 and use the remainder:

• $i^7$: $7 \div 4 = 1$ remainder $3$, so $i^7 = i^3 = -i$
• $i^{15}$: $15 \div 4 = 3$ remainder $3$, so $i^{15} = i^3 = -i$
• $i^{20}$: $20 \div 4 = 5$ remainder $0$, so $i^{20} = i^4 = 1$

Forms of Complex Numbers

Algebraic form (rectangular form) is the standard $a + bi$ you've been using. Examples: $3 + 2i$, $-4 - 7i$.

Polar form expresses a complex number using its distance from the origin and its angle. It's written as $r(\cos\theta + i\sin\theta)$, where:

• Modulus $r = \sqrt{a^2 + b^2}$ (the distance from the origin)
• Argument $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ (the angle measured from the positive real axis)

For example, converting $2 + 2i$ to polar form:

1. Find the modulus: $r = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$
2. Find the argument: $\theta = \tan^{-1}\left(\frac{2}{2}\right) = 45°$ or $\frac{\pi}{4}$ radians
3. Write it: $2\sqrt{2}\left(\cos 45° + i\sin 45°\right)$

Converting from polar back to algebraic uses $a = r\cos\theta$ and $b = r\sin\theta$. For example:

$2(\cos 30° + i\sin 30°) = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \sqrt{3} + i$

Graphing on the Complex Plane

The complex plane looks like a regular coordinate plane, but the horizontal axis represents the real part and the vertical axis represents the imaginary part. Each complex number $a + bi$ corresponds to the point $(a, b)$.

• $3 - 4i$ is plotted 3 units right and 4 units down from the origin.
• The modulus $r$ is the point's distance from the origin.
• The argument $\theta$ is the angle the point makes with the positive real axis.

This geometric view is what makes polar form intuitive: $r$ tells you how far from the origin, and $\theta$ tells you in what direction.

De Moivre's Theorem

De Moivre's theorem provides a shortcut for raising complex numbers to a power when they're in polar form:

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

You raise the modulus to the $n$th power and multiply the argument by $n$. This is far easier than multiplying a complex number by itself repeatedly in algebraic form. Your course may not go deep into this, but recognizing the formula is useful if it appears on an exam.