x2=−1x² = −1 has no real solution. Is inventing one just cheating?

The number $i$i and Cartesian form

• Define $i$i by i2=−1i² = −1. Then square roots of negatives exist: −16​=16​i=4i√(−16) = √16 i = 4i.
• A complex number in Cartesian form is $z=a+bi$z = a + bi with $a,b\in \mathbb{R}$a, b ∈ ℝ.
• $a$a is the real part, $\mathrm{Re}(z)=a$Re(z) = a. $b$b is the imaginary part, $\mathrm{Im}(z)=b$Im(z) = b: a real number, with no $i$i.
• Real numbers have $b=0$b = 0; purely imaginary numbers have $a=0$a = 0.
• Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.

Arithmetic and the conjugate

• Add or subtract: combine real parts and imaginary parts separately.
• Multiply: expand the brackets as usual, then replace i2i² with $-1$−1.
• The complex conjugate of $z=a+bi$z = a + bi is z∗=a−biz^(*) = a − bi. Then zz∗=a2+b2zz^(*) = a² + b², which is real.
• Divide: multiply the top and bottom by the conjugate of the denominator, so the denominator becomes real.

The Argand diagram and modulus

• On an Argand diagram (the complex plane), $z=a+bi$z = a + bi is the point $(a,b)$(a, b): real axis across, imaginary axis up.
• The modulus ∣z∣=a2+b2​|z| = √(a² + b²) is the distance from the origin to the point. For example, $|3+4i|=5$|3 + 4i| = 5.
• The argument $\arg z$arg z is the angle from the positive real axis to the point (used in the next mission).
• z∗z^(*) is the reflection of $z$z in the real axis.
• Adding complex numbers adds them like vectors, so ∣z1​−z2​∣|z₁ − z₂| is the distance between the points z1​z₁ and z2​z₂.

Why does multiplying by $i$i rotate a point by 90°?

Modulus–argument form

• Any $z\ne 0$z ≠ 0 is fixed by its modulus $r=|z|$r = |z| and argument $\theta =\arg z$θ = arg z, the angle from the positive real axis.
• Polar form: $z=r(\cos \theta +i\sin \theta )=r\mathrm{cis}\theta$z = r(cos θ + isin θ) = r cis θ. Euler form (also called exponential form): z=reiθz = re^iθ.
• Converting: $a=r\cos \theta$a = rcos θ, $b=r\sin \theta$b = rsin θ, and r=a2+b2​r = √(a² + b²).
• Find $\theta$θ from a sketch: work out the acute angle from tanα=​ab​​tan α = |b/a|, then place it in the right quadrant. The principal argument is $-\pi <\theta \le \pi$−π < θ ≤ π.

Multiplying and dividing in Euler form

• Multiply: r1​eiθ1​×r2​eiθ2​=r1​r2​ei(θ1​+θ2​)r₁e^iθ₁ × r₂e^iθ₂ = r₁ r₂ e^(i(θ₁ + θ₂)): multiply the moduli, add the arguments.
• Divide: r2​eiθ2​r1​eiθ1​​=r2​r1​​ei(θ1​−θ2​)(r₁e^iθ₁)/(r₂e^iθ₂) = r₁/r₂ e^(i(θ₁ − θ₂)): divide the moduli, subtract the arguments.
• Geometrically, multiplying by $w$w stretches by $|w|$|w| and rotates by $\arg w$arg w about the origin.
• Adjust the final argument by $\pm 2\pi$± 2π to bring it into $-\pi <\theta \le \pi$−π < θ ≤ π.

$n$nth roots of a complex number

• De Moivre's theorem also works for rational powers, which gives roots.
• Write the number with every possible argument: w=rei(θ+2kπ)w = re^(i(θ + 2kπ)), $k\in \mathbb{Z}$k ∈ ℤ.
• The solutions of zn=wzⁿ = w are z=rn1​einθ+2kπ​z = r^(1/n) e^(i(θ + 2kπ)/n) for $k=0,1,\dots ,n-1$k = 0, 1, …, n − 1: exactly $n$n roots.
• The roots all have modulus rn1​r^(1/n) and are spaced n2π​2π/n apart: they form a regular $n$n-gon centred at the origin.

A real cubic has root $2+i$2 + i. What other root is guaranteed?

Conjugate roots of real polynomials

• If $p(z)$p(z) has real coefficients and $p(w)=0$p(w) = 0, then p(w∗)=0p(w^(*)) = 0: non-real roots come in conjugate pairs.
• Why: the conjugate of a sum is the sum of conjugates, and of a product the product of conjugates. Real coefficients are unchanged by conjugating, so (p(w))∗=p(w∗)(p(w))^(*) = p(w^(*)), and 0∗=00^(*) = 0.
• The pair $w=a\pm bi$w = a ± bi gives a real quadratic factor: (z−w)(z−w∗)=z2−2az+(a2+b2)(z − w)(z − w^(*)) = z² − 2az + (a² + b²).
• So a real cubic always has at least one real root.

De Moivre's theorem by induction

• De Moivre's theorem: (cosθ+isinθ)n=cosnθ+isinnθ(cos θ + isin θ)ⁿ = cos nθ + isin nθ.
• Proof for n∈Z+n ∈ ℤ⁺ by induction: true for $n=1$n = 1. Assume true for $n=k$n = k, then multiply both sides by $\cos \theta +i\sin \theta$cos θ + isin θ.
• Expand and use compound angle identities to reach $\cos (k+1)\theta +i\sin (k+1)\theta$cos (k+1)θ + isin (k+1)θ, so true for $n=k+1$n = k + 1.
• Conclude: true for $n=1$n = 1, and true for $k$k implies true for $k+1$k + 1, so true for all n∈Z+n ∈ ℤ⁺.
• It also holds for negative integers $n$n, and with rational $n$n it gives roots: coskθ​+isinkθ​cos θ/k + isin θ/k is one of the $k$kth roots of $\cos \theta +i\sin \theta$cos θ + isin θ.

Multiple-angle identities

• Write $c=\cos \theta$c = cos θ, $s=\sin \theta$s = sin θ. By De Moivre, cos3θ+isin3θ=(c+is)3cos 3θ + isin 3θ = (c + is)³.
• Expand with the binomial theorem: (c+is)3=c3+3ic2s−3cs2−is3(c + is)³ = c³ + 3ic²s − 3cs² − is³.
• Equate real parts: cos3θ=c3−3cs2cos 3θ = c³ − 3cs². With s2=1−c2s² = 1 − c², this gives cos3θ=4cos3θ−3cosθcos 3θ = 4cos³θ − 3cos θ.
• Equate imaginary parts to get $\sin 3\theta$sin 3θ.

Big ideas

• $z=a+bi$z = a + bi with i2=−1i² = −1: $\mathrm{Re}(z)=a$Re(z) = a, $\mathrm{Im}(z)=b$Im(z) = b, conjugate z∗=a−biz^(*) = a − bi, modulus ∣z∣=a2+b2​|z| = √(a² + b²).
• Multiply by expanding and using i2=−1i² = −1; divide by multiplying top and bottom by the conjugate of the denominator.
• On the Argand diagram $z$z is the point $(a,b)$(a, b); a real quadratic with b2−4ac<0b² − 4ac < 0 has a conjugate pair of complex roots.
• z=r(cosθ+isinθ)=rcisθ=reiθz = r(cos θ + isin θ) = r cis θ = re^iθ: find $r$r by Pythagoras and $\theta$θ from a sketch, with $-\pi <\theta \le \pi$−π < θ ≤ π.
• Multiply moduli and add arguments; divide moduli and subtract arguments; for powers, (reiθ)n=rneinθ(re^iθ)ⁿ = rⁿe^inθ. Multiplying by $w$w is a stretch by $|w|$|w| and a rotation by $\arg w$arg w.
• The $n$n roots of zn=wzⁿ = w have modulus rn1​r^(1/n) and arguments nθ+2kπ​(θ + 2kπ)/n: a regular $n$n-gon.
• A polynomial with real coefficients has its non-real roots in conjugate pairs; $a\pm bi$a ± bi gives the real factor z2−2az+a2+b2z² − 2az + a² + b².
• De Moivre's theorem, (cosθ+isinθ)n=cosnθ+isinnθ(cos θ + isin θ)ⁿ = cos nθ + isin nθ, is proved for positive integers by induction using compound angle identities.
• Expand (c+is)n(c + is)ⁿ and equate real or imaginary parts to get identities for $\cos n\theta$cos nθ and $\sin n\theta$sin nθ.