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MHT-CET Maths · Formula sheet

Complex Numbers formulas

15 formulas and 15 common traps for MHT-CET Maths Complex Numbers, grouped by subtopic.

Full notes with worked examples

Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity

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Powers of i and the Standard Form x + iy

Cycle of powers of i

i2=−1,i3=−i,i4=1,in=i n mod 4i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,\qquad i^n = i^{\,n \bmod 4}

The Conjugate: Rationalising a Denominator and Equating Conjugates

Conjugate identities

zzˉ=∣z∣2z1z2=z1zˉ2∣z2∣2z1=zˉ2  ⟺  Re⁡z1=Re⁡z2, Im⁡z1=−Im⁡z2z\bar z = |z|^2 \qquad \frac{z_1}{z_2} = \frac{z_1\bar z_2}{|z_2|^2} \qquad z_1 = \bar z_2 \iff \operatorname{Re}z_1 = \operatorname{Re}z_2,\ \operatorname{Im}z_1 = -\operatorname{Im}z_2

Purely Real or Purely Imaginary: Set the Other Part to Zero

Real and imaginary conditions

z purely imaginary  ⟺  Re⁡z=0z purely real  ⟺  Im⁡z=0sin⁡2θ=sin⁡2α⇒θ=nπ±αz \text{ purely imaginary} \iff \operatorname{Re}z = 0 \qquad z \text{ purely real} \iff \operatorname{Im}z = 0 \qquad \sin^2\theta = \sin^2\alpha \Rightarrow \theta = n\pi \pm \alpha

Solving for z: Put z = x + iy and Equate Parts

Equating parts

a+ib=c+id  ⟺  a=c and b=d11−i=1+i2a + ib = c + id \iff a = c \text{ and } b = d \qquad \frac{1}{1 - i} = \frac{1 + i}{2}

A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions

Minimal quadratic and the cube

x=a+ib⇒x2−2ax+(a2+b2)=0(p+iq)3=(p3−3pq2)+i(3p2q−q3)x = a + ib \Rightarrow x^2 - 2ax + (a^2 + b^2) = 0 \qquad (p + iq)^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)

Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0

The two facts

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

Common traps

i³ = i, and other lapses in the cycle

i3=i2⋅i=−ii^3 = i^2\cdot i = -i. Reading i35i^{35} as ii instead of −i-i flips a sign in the imaginary part and lands on the wrong modulus, which is always offered.

Conjugating only the numerator

Z1‾/Z2‾\overline{Z_1}/\overline{Z_2} needs BOTH conjugates before the division; and the division itself still needs a second conjugate multiplication to clear the denominator. Two conjugations, not one.

Setting the imaginary part to zero for 'purely imaginary'

Purely imaginary means NO real part. The condition is Re⁡=0\operatorname{Re} = 0; solving Im⁡=0\operatorname{Im} = 0 gives θ=nπ\theta = n\pi, which is offered as option (D).

Dividing by 1 − i without rationalising

in2(1−i)\dfrac{in}{2(1 - i)} is not yet in standard form; multiply by 1+i1+i\dfrac{1 + i}{1 + i} first. Reading the real part off the un-rationalised form gives Re⁡z=0\operatorname{Re}z = 0 and a wrong sign on nn.

Substituting the complex number directly

Raising 1+2i1 + 2i to the fourth power by hand invites a sign error at every step. The minimal quadratic reduces the polynomial to its remainder in two lines of long division.

Stopping at ω⁴

16ω416\omega^4 is not an option; 16ω16\omega is. Reduce every exponent modulo 33 before matching, and remember ω2\omega^2 is a different option from ω\omega.

Modulus and Argument — Polar Form, De Moivre and Square Roots

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Modulus and Its Properties: |z₁z₂| = |z₁||z₂|

Modulus rules

∣z1z2∣=∣z1∣∣z2∣∣z1z2∣=∣z1∣∣z2∣∣zn∣=∣z∣n∣z∣=∣z∣|z_1z_2| = |z_1||z_2| \qquad \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad |z^n| = |z|^n \qquad |\sqrt z| = \sqrt{|z|}

The Argument: Reference Angle Plus the Quadrant

Argument rules

arg⁡(z1z2)=arg⁡z1+arg⁡z2arg⁡z1z2=arg⁡z1−arg⁡z2arg⁡zˉ=−arg⁡z\arg(z_1z_2) = \arg z_1 + \arg z_2 \qquad \arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2 \qquad \arg\bar z = -\arg z

Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ

De Moivre and the rotation by i

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθiz is z rotated by π2sin⁡θ+icos⁡θ=i(cos⁡θ−isin⁡θ)(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \qquad iz \text{ is } z \text{ rotated by } \tfrac{\pi}{2} \qquad \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta)

|z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z|

Closed form

∣z∣+z=a+ib ⇒ Im⁡z=b,∣z∣=a2+b22a|z| + z = a + ib \ \Rightarrow\ \operatorname{Im}z = b,\quad |z| = \frac{a^2 + b^2}{2a}

Find z From a Given Modulus: Simplify, Then Fix the Parameter

Modulus first, algebra second

∣(1+i)2a−i∣=2a2+1∣2+cos⁡θ+isin⁡θ∣2=5+4cos⁡θ\left|\frac{(1 + i)^2}{a - i}\right| = \frac{2}{\sqrt{a^2 + 1}} \qquad |2 + \cos\theta + i\sin\theta|^2 = 5 + 4\cos\theta

Common traps

Expanding the product to find its modulus

(3+i)3(4+3i)2(\sqrt3 + i)^3(4 + 3i)^2 expanded is a page of algebra with several sign traps. Moduli multiply: 23⋅522^3\cdot5^2. Never expand for a modulus.

The argument from the ratio alone

tan⁡−13/2−1/2=tan⁡−1(−3)=−π3\tan^{-1}\dfrac{\sqrt3/2}{-1/2} = \tan^{-1}(-\sqrt3) = -\dfrac{\pi}{3} is the fourth-quadrant answer for a second-quadrant point. Plot the point first; the un-adjusted angle is always an option.

Applying De Moivre to sin θ + i cos θ

(sin⁡θ+icos⁡θ)5(\sin\theta + i\cos\theta)^5 is NOT sin⁡5θ+icos⁡5θ\sin 5\theta + i\cos 5\theta. Rewrite it as i(cos⁡θ−isin⁡θ)i(\cos\theta - i\sin\theta) first; the i5=ii^5 = i in front is where the final sin⁡9θ−icos⁡9θ\sin 9\theta - i\cos 9\theta comes from.

Answering the twin sitting's value

54\frac54 belongs to 2+i2 + i and 53\frac53 to 3+i3 + i; each list offers both. Read the right-hand side before recalling the number.

z or z̄?

The 2023 sitting asks for zˉ\bar z and the 2024 sitting for zz, with sign-flipped options in each. Compute zz, then read the last word of the stem before choosing.

Locus in the Argand Plane — Circles, Lines and Greatest/Least Modulus

Learn this subtopic in the notes

|z − a| = r Is a Circle: Centre a, Radius r

Circle in the Argand plane

∣z−a∣=r  ⟺  circle, centre a, radius r∣z−az−b∣=k≠1  ⟺  circle (Apollonius)|z - a| = r \iff \text{circle, centre } a,\ \text{radius } r \qquad \left|\frac{z - a}{z - b}\right| = k \ne 1 \iff \text{circle (Apollonius)}

|z − a| = |z − b| Is the Perpendicular Bisector of ab

Line from equal distances

∣z−a∣=∣z−b∣  ⟺  perpendicular bisector of a,b∣z−a∣−∣z−b∣=∣a−b∣  is a ray (degenerate hyperbola)|z - a| = |z - b| \iff \text{perpendicular bisector of } a, b \qquad |z - a| - |z - b| = |a - b| \ \text{ is a ray (degenerate hyperbola)}

Re of a Quotient Equals Zero: Rationalise, Then Read the Circle

Purely imaginary quotient

Re⁡z−12z+1=0  ⟺  2x2+2y2−x−1=0  ⟺  (x−14)2+y2=916\operatorname{Re}\frac{z - 1}{2z + 1} = 0 \iff 2x^2 + 2y^2 - x - 1 = 0 \iff \left(x - \tfrac14\right)^2 + y^2 = \tfrac{9}{16}

Greatest and Least |z| on a Disc: |a| + r and |a| − r

Extreme modulus on a disc

∣z−a∣≤r:∣a∣−r≤∣z∣≤∣a∣+rmax⁡∣z∣−min⁡∣z∣=2r (∣a∣≥r)|z - a| \le r:\quad |a| - r \le |z| \le |a| + r \qquad \max|z| - \min|z| = 2r \ (|a| \ge r)

Common traps

Reading |z + 1| as centred at +1

∣z+1∣=∣z−(−1)∣|z + 1| = |z - (-1)|: the centre is −1-1. The option 'centre (1,0)(1, 0)' is built for this sign slip.

Calling every modulus locus a circle

When the two moduli have equal weight the squares cancel and the locus is a LINE. 'Circle' is the first option on every such stem for the student who did not square.

Reporting the radius squared

(x−14)2+y2=916\left(x - \frac14\right)^2 + y^2 = \frac{9}{16} has radius 34\frac34; 916\frac{9}{16} is the first distractor on the list.

Answering 2√5 for the difference

The difference of the extreme moduli is 2r2r, the diameter, not 2∣a∣2|a|. For ∣z−2+i∣≤2|z - 2 + i| \le 2 that is 44; 252\sqrt5 is the planted wrong answer.

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