i2 = −1
$\therefore i = \sqrt {-1}$
$\sqrt{-2} = \sqrt{-1 \times 2}$
$= \sqrt{2}i$
ℂ = {a + bi : a, b ∈ ℝ}
General form: z = a + bi - Re(z)=a - Im(z)=b
If z1 = a + bi and z2 = c + di, then
z1 + z2 = (a + c)+(b + d)i
If z1 = a + bi and z2 = c + di, then
z1 − z2 = (a − c)+(b − d)i
If z = a + bi and k ∈ ℝ, then
kz = ka + kbi
i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 …
Therefore.. - i4n = 1 - i4n + 1 = i - i4n + 2 = −1 - i4n + 3 = −i
If z1 = a + bi and z2 = c + di, then z1 × z2 = (ac − bd)+(ad + bc)i
If z = a + bi, conjugate of z is $\overline{z} = a-bi$ (flipped operator)
Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2$
Distance from origin. $|{z}|=\sqrt{a^2+b^2}$
$\therefore z \overline{z} = |z|^2$
$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$
${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$
(using multiplicative inverse)
In practice, rationalise denominator: ${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
To solve z2 + a2 = 0 (sum of two squares):
z2 + a2 = z2 − (ai)2 =(z + ai)(z − ai)
General form: z = rcisθ =rcosθ + rsinθi
where - z = a + bi - r is the distance from origin, given by Pythagoras ($r=\sqrt{x^2+y^2}$) - θ is the argument of z, CCW from origin
Note each complex number has multiple polar representations: z = rcisθ = rcis(θ + 2nπ) where n is integer number of revolutions
z1z2 = r1r2cis(θ1 + θ2) (multiply moduli, add angles)
${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
(rcisθ)n = rncis(nθ)