Complex Arithmetic

Roots of Negative Numbers #

No real number squares to a negative.

No real number squares to a negative, so $x^2=-1$x² = −1 has no solution on the number line

Squaring a real number never gives a negative. The curve never dips below 0.

So $y=x^2$y = x² never meets y = −1, and $x^2=-1$x² = −1 has no real solution.

That is why $x^2+1=0$x² + 1 = 0 has no real solution yet, and why this curve misses the x-axis.

The Imaginary Unit #

Define one new number whose square is minus one.

Define one new number whose square is minus one and every other root follows

Multiplying by i is a quarter turn. Start at 1 and you land on i.

Turn again and you land on −1. Two quarter turns take 1 to −1, so $i^2=-1$i² = −1.

Every root of a negative now has an answer: $\sqrt{-9}$√(−9) is $\sqrt{9}\times \sqrt{-1}$√9 × √(−1), which is 3i.

The real part goes across and the imaginary part goes up. Together they make a complex number.

Powers of i #

Keep turning: the powers repeat every four.

Multiplying by i keeps turning, so the powers of i repeat in a cycle of four

Keep turning: a quarter turn from −1 lands on −i. So $i^3=-i$i³ = −i.

One more turn closes the circle: $i^4=1$i⁴ = 1, back where it started.

The powers repeat in a cycle of four — only the remainder of $n/4$n / 4 matters.

Split off the full turns: $i^4=1$i⁴ = 1, so only the remainder decides the value.

Adding Complex Numbers #

Add the real parts, then the imaginary parts.

Add complex numbers by adding the real parts and the imaginary parts separately

Start the 1 + 4i arrow where the 3 + 2i arrow ends. You land on 4 + 6i.

Across 3 then 1 makes 4; up 2 then 4 makes 6. The two parts never mix.

Subtracting reverses the second arrow: (5 + 3i) − (2 + i) = 3 + 2i.

Adding complex numbers is collecting like terms, with i in the place of a letter.

Multiplying Complex Numbers #

Multiply out as usual, then replace i squared.

Multiply out as usual, then replace i squared with minus one

Expand as if i were a letter, then use the one rule: $i^2=-1$i² = −1.

Check it on one case: the same expansion turns (1 + i)(1 + i) into 2i.

Read the product as arrows: $\sqrt{2}\times \sqrt{2}=2$√2 × √2 = 2 and 45° + 45° = 90°. Lengths multiply, angles add.

The rule holds every time: 2 at 30° times 3 at 60° is 6 at 90°.

Complex Conjugates #

Flip the sign of the imaginary part.

Flipping the sign of the imaginary part gives the conjugate, and the product of the pair is real

The conjugate keeps the real part and flips the sign of the i part.

The pair is a reflection in the real axis. Multiplying them gives a difference of two squares, $a^2-b^2$a² − b².

Multiply the pair out: the middle terms cancel and $-4i^2$−4i² turns into +4.

The answer 13 lands on the real axis, with no imaginary part left at all.

Dividing Complex Numbers #

Multiply top and bottom by the conjugate.

Multiply top and bottom by the conjugate to clear i out of the denominator

A complex number in the denominator is not a finished answer, just as a radical is not.

Multiply numerator and denominator by the conjugate: that is multiplying by 1.

The denominator is (2 + i)(2 − i) = 5, a real number, so the division is ordinary: $\frac{\left(2-i\right)}{5}=0.4-0.2i$(2 − i)/5 = 0.4 − 0.2i.

Complex Roots of a Quadratic #

A quadratic that misses the axis still has roots.

A quadratic that misses the axis still has two roots, and they are conjugates

The curve misses the axis: the discriminant is negative, so its square root brings in i.

Solve the equation of the drawn curve: $x^2+1=0$x² + 1 = 0 gives $x^2=-1$x² = −1, so x is $\pm i$±i.

So there are two roots after all — a conjugate pair, when the coefficients are real.

Complex Roots of a Cubic #

One non-real root brings its conjugate along.

A real cubic with one non-real root carries its conjugate too, leaving one real root

You are given a cubic with real coefficients and one of its non-real roots.

The conjugate rule gives a second root at once: 2 − 3i must be a root too.

Multiply that pair of factors and the i vanishes: $z^2-4z+13$z² − 4z + 13, all real.

Divide the cubic by that real quadratic. What is left is the linear factor z − 1.

So the roots are 2 + 3i, its conjugate 2 − 3i, and the real root 1.