PYQ Vault

NDA Mathematics · Formula sheet

Complex Numbers formulas

8 formulas and 8 common traps for NDA Mathematics Complex Numbers, grouped by subtopic.

Full notes with worked examples

Modulus, Argument & Conjugate

Learn this subtopic in the notes

What a complex number is

Fundamentals of a complex number

i2=−1a+ib=c+id  ⟺  a=c, b=d(a+ib)(c+id)=(ac−bd)+i(ad+bc)i^2=-1 \qquad a+ib=c+id \iff a=c,\ b=d \qquad (a+ib)(c+id)=(ac-bd)+i(ad+bc)

Conjugate; purely real / purely imaginary

Conjugate identities

a+ib‾=a−ibzzˉ=∣z∣2Re⁡(z)=z+zˉ2Im⁡(z)=z−zˉ2iz1z2‾=zˉ1 zˉ2\overline{a+ib}=a-ib \qquad z\bar z=|z|^2 \qquad \operatorname{Re}(z)=\dfrac{z+\bar z}{2} \qquad \operatorname{Im}(z)=\dfrac{z-\bar z}{2i} \qquad \overline{z_1z_2}=\bar z_1\,\bar z_2

Modulus and the triangle inequality

Modulus properties

∣z∣=a2+b2∣z1z2∣=∣z1∣ ∣z2∣∣z1z2∣=∣z1∣∣z2∣∣z∣2=zzˉ∣z1+z2∣≤∣z1∣+∣z2∣|z|=\sqrt{a^2+b^2} \qquad |z_1z_2|=|z_1|\,|z_2| \qquad \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|} \qquad |z|^2=z\bar z \qquad |z_1+z_2|\le|z_1|+|z_2|

Argument and polar form

Polar form and argument

z=r(cos⁡θ+isin⁡θ)=reiθarg⁡(z1z2)=arg⁡z1+arg⁡z2arg⁡ ⁣(z1z2)=arg⁡z1−arg⁡z2z=r(\cos\theta+i\sin\theta)=re^{i\theta} \qquad \arg(z_1z_2)=\arg z_1+\arg z_2 \qquad \arg\!\left(\dfrac{z_1}{z_2}\right)=\arg z_1-\arg z_2

Loci in the Argand plane — which curve is it?

Translation dictionary

zzˉ=x2+y2,z+zˉ=2x,z−zˉ=2iy,∣z−a∣=r (circle),∣z−a∣=∣z−b∣ (perpendicular bisector)z\bar z=x^2+y^2,\quad z+\bar z=2x,\quad z-\bar z=2iy,\quad |z-a|=r\ \text{(circle)},\quad |z-a|=|z-b|\ \text{(perpendicular bisector)}

Common traps

Purely imaginary is the real-part-zero condition, not the imaginary-part-zero one

A number is purely imaginary when z=−zˉz=-\bar z (its real part is 0), and purely real when z=zˉz=\bar z (its imaginary part is 0). Students routinely swap these. For z=(x−2)+3iz=(x-2)+3i to be purely imaginary you set the real part x−2=0x-2=0, not the imaginary part.

zzˉ=∣z∣2z\bar z=|z|^2, not ∣z∣|z|

The product zzˉz\bar z equals the modulus squared: zzˉ=a2+b2=∣z∣2z\bar z=a^2+b^2=|z|^2. Forgetting the square (writing zzˉ=∣z∣z\bar z=|z|) is the single most common modulus slip. So for z=3+4iz=3+4i, zzˉ=25z\bar z=25, while ∣z∣=5|z|=5.

Modulus does not distribute over a sum

∣z1+z2∣≠∣z1∣+∣z2∣|z_1+z_2|\ne|z_1|+|z_2| in general — that's only an inequality (∣z1+z2∣≤∣z1∣+∣z2∣|z_1+z_2|\le|z_1|+|z_2|), with equality only when z1,z2z_1,z_2 point the same way. Modulus DOES distribute over products and quotients: ∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|. For maxima/minima of ∣z±c∣|z\pm c|, reach for the triangle inequality, never term-by-term addition.

The principal argument depends on the quadrant, not just tan⁡−1(b/a)\tan^{-1}(b/a)

tan⁡−1(b/a)\tan^{-1}(b/a) alone can't tell apart a+iba+ib from −a−ib-a-ib (same ratio, opposite quadrants). Find the reference angle tan⁡−1∣ba∣\tan^{-1}\big|\tfrac{b}{a}\big|, then place it by the signs of (a,b)(a,b) so the result lands in (−π,π](-\pi,\pi]. For −1+i-1+i (2nd quadrant) the argument is 3π4\tfrac{3\pi}4, not tan⁡−1(−1)=−π4\tan^{-1}(-1)=-\tfrac\pi4.

A third-quadrant answer may be keyed in [0,2π)[0,2\pi)

1−i31+i3=e−2πi/3\dfrac{1-i\sqrt3}{1+i\sqrt3} = e^{-2\pi i/3} has principal argument −2π3-\tfrac{2\pi}{3} in (−π,π](-\pi,\pi] — but the paper keyed it 240∘240^\circ, the [0,2π)[0,2\pi) reading. Neither is a mistake; they differ by 2π2\pi. Compute in your convention, then match the option list — do not reject a correct angle because it is written on the other branch.

∣z−a∣=∣z−b∣|z-a|=|z-b| is a LINE, not a circle

Equal distances from two fixed points is the perpendicular bisector. A single ∣z−a∣=r|z-a|=r is a circle; the ratio form ∣z−a∣=k∣z−b∣|z-a|=k|z-b| is a circle only for k≠1k\ne1. Students see two moduli and answer 'circle'.

Powers of i, De Moivre & Roots

Learn this subtopic in the notes

Powers of i (the period-4 cycle)

Powers of i

i2=−1i3=−ii4=1i4k+r=irik+ik+1+ik+2+ik+3=0i^2=-1 \qquad i^3=-i \qquad i^4=1 \qquad i^{4k+r}=i^r \qquad i^k+i^{k+1}+i^{k+2}+i^{k+3}=0

De Moivre's theorem and roots

De Moivre's theorem and nth roots

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθzn=rneinθz1/n=r1/nei(θ+2kπ)/n, k=0,…,n−1(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta \qquad z^n=r^n e^{in\theta} \qquad z^{1/n}=r^{1/n}e^{i(\theta+2k\pi)/n},\ k=0,\ldots,n-1

Common traps

Reduce the exponent of ii mod 4 — and watch the remainder

in=i n mod 4i^n=i^{\,n\bmod 4}: take the exponent mod 4, not the whole number mod something else. Remainder 0→10\to1, 1→i1\to i, 2→−12\to-1, 3→−i3\to-i. So i102i^{102}: 102 mod 4=2⇒i102=−1102\bmod4=2\Rightarrow i^{102}=-1 (a frequent error is reading remainder 2 as ii instead of −1-1).

Cube Roots of Unity

Learn this subtopic in the notes

1, ω, ω² and their identities

Cube roots of unity identities

ω3=11+ω+ω2=0ωˉ=ω2ωn=ω n mod 3\omega^3=1 \qquad 1+\omega+\omega^2=0 \qquad \bar\omega=\omega^2 \qquad \omega^n=\omega^{\,n\bmod 3}

Common traps

ω2=ωˉ\omega^2=\bar\omega, but ω2≠−ω\omega^2\ne-\omega

ω2\omega^2 is the conjugate ωˉ\bar\omega (both unit-circle cube roots, 120° apart). From 1+ω+ω2=01+\omega+\omega^2=0 you get ω2=−1−ω\omega^2=-1-\omega — not −ω-\omega. Treating ω2\omega^2 as −ω-\omega (or forgetting to reduce ωn\omega^n by n mod 3n\bmod 3 first) wrecks the algebra. Also note ω⋅ω2=ω3=1\omega\cdot\omega^2=\omega^3=1.

More NDA Mathematics formula sheets