Complex Numbers

Why we invent $i$

No real number squared gives $-1$. So we invent one and call it $i$:

$i^2 = -1 \qquad i = \sqrt{-1}$

A complex number has the form $a + bi$, where $a$ and $b$ are real.

• $a$ is the real part.
• $b$ is the imaginary part.

Examples: $3 + 2i$, $-5 - i$, and $7$ (which is $7 + 0i$ — every real number is also complex).

Powers of $i$ cycle

$i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,\quad i^5 = i,\ \ldots$

To compute $i^n$, divide $n$ by $4$ and use the remainder.

Example. $i^{27}$: $27 \div 4 = 6$ remainder $3$, so $i^{27} = i^3 = -i$.

Arithmetic

Add/subtract by combining real with real and imaginary with imaginary.

$(3 + 2i) + (1 - 5i) = 4 - 3i$

Multiply like binomials, then replace $i^2$ with $-1$.

$(2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i^2 = 2 - 5i + 12 = 14 - 5i$

Complex conjugates

The conjugate of $a + bi$ is $a - bi$. Multiplying a complex number by its conjugate gives a real result:

$(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2$

That's how you divide complex numbers — multiply top and bottom by the conjugate of the bottom.

Example. $\dfrac{3 + i}{2 - i}$. Multiply by $\dfrac{2 + i}{2 + i}$:

$\dfrac{(3 + i)(2 + i)}{(2 - i)(2 + i)} = \dfrac{6 + 3i + 2i + i^2}{4 + 1} = \dfrac{5 + 5i}{5} = 1 + i$

Quadratics with complex roots

When the discriminant $b^2 - 4ac$ is negative, the quadratic formula gives complex roots.

Example. $x^2 - 2x + 5 = 0$. Discriminant: $4 - 20 = -16$.

$x = \dfrac{2 \pm \sqrt{-16}}{2} = \dfrac{2 \pm 4i}{2} = 1 \pm 2i$

For real-coefficient polynomials, complex roots always come in conjugate pairs.

Key takeaways

• $i^2 = -1$. A complex number $a + bi$ has a real and imaginary part.
• Powers of $i$ cycle through $i, -1, -i, 1$.
• Multiply complex numbers like binomials; replace $i^2$ with $-1$.
• Divide by multiplying top and bottom by the conjugate of the denominator.
• Quadratics with negative discriminant have a conjugate pair of complex roots.