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

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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

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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

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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

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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

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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

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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, $a^2-{\left(\text{bi}\right)}^2=a^2+b^2$a² − (bi)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

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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: $\frac{2-i}{5}=0.4-0.2i$(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

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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

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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.