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EAPCET Mathematics · Intermediate second year · Paper IIA

Complex Numbers

Complex Numbers covers the algebra of numbers of the form a + ib, their modulus and argument, and their geometry in the Argand plane.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • i4n=1i^{4n} = 1, i4n+1=ii^{4n+1} = i, i4n+2=−1i^{4n+2} = -1, i4n+3=−ii^{4n+3} = -i
  • zzˉ=∣z∣2z\bar{z} = |z|^{2}
  • ∣z1z2∣=∣z1∣∣z2∣|z_{1}z_{2}| = |z_{1}||z_{2}| and arg⁡(z1z2)=arg⁡z1+arg⁡z2\arg(z_{1}z_{2}) = \arg z_{1} + \arg z_{2}
  • Triangle inequality: ∣z1+z2∣≤∣z1∣+∣z2∣|z_{1} + z_{2}| \le |z_{1}| + |z_{2}|
  • ∣z−z0∣=r|z - z_{0}| = r is a circle with centre z0z_{0} and radius rr
  • Principal argument of z=x+iyz = x + iy with α=tan⁡−1∣yx∣\alpha = \tan^{-1}\left|\tfrac{y}{x}\right|: α\alpha in quadrant I, π−α\pi - \alpha in II, −(π−α)-(\pi - \alpha) in III, −α-\alpha in IV
  • arg⁡(zˉ)=−arg⁡z\arg(\bar{z}) = -\arg z and ∣zˉ∣=∣z∣|\bar{z}| = |z| (for zz not a negative real number)
  • arg⁡(z1z2)=arg⁡z1−arg⁡z2\arg\left(\dfrac{z_{1}}{z_{2}}\right) = \arg z_{1} - \arg z_{2} and arg⁡(zn)=narg⁡z\arg(z^{n}) = n\arg z (up to multiples of 2π2\pi)
  • Euler form: cos⁡θ+isin⁡θ=eiθ\cos\theta + i\sin\theta = e^{i\theta}, so z=∣z∣eiθz = |z|e^{i\theta}
  • ii=e−π/2i^{i} = e^{-\pi/2} (principal value)
  • ∣∣z1∣−∣z2∣∣≤∣z1±z2∣≤∣z1∣+∣z2∣\big||z_{1}| - |z_{2}|\big| \le |z_{1} \pm z_{2}| \le |z_{1}| + |z_{2}|

What this chapter covers

  • Algebra of complex numbers
  • Conjugate, modulus and argument
  • Polar form
  • Argand plane and loci
  • Square roots of complex numbers

How it is usually asked

  • Find a modulus or argument
  • Find the locus of z from a condition on |z|
  • Simplify an expression to the form a + ib

Understanding Complex Numbers

A complex number z = a + ib has a real part a and an imaginary part b, where i² = −1. Complex numbers are added and multiplied like algebraic expressions, replacing i² by −1 wherever it appears, and powers of i repeat in a cycle of four. To divide, multiply the numerator and denominator by the conjugate of the denominator, z̄ = a − ib, which makes the denominator real.

Each complex number is also a point (a, b) in the Argand plane. Its distance from the origin is the modulus |z| = √(a² + b²), and the angle its position makes with the positive real axis is the argument. The principal argument lies in (−π, π], and its quadrant must be read from the signs of a and b. Moduli multiply and arguments add when complex numbers are multiplied.

The geometric view turns conditions on z into curves. |z − z₀| is the distance from z to z₀, so |z − z₀| = r is a circle, and |z − z₁| = |z − z₂| is the perpendicular bisector of the segment joining z₁ and z₂. EAPCET questions ask you to simplify to the form a + ib, find a modulus or argument, or identify such a locus.

Worked example

EAPCET-style question

The principal argument of 7+i3+4i\dfrac{7 + i}{3 + 4i} is

  1. (1)π4\dfrac{\pi}{4}
  2. (2)−π4-\dfrac{\pi}{4}
  3. (3)3π4\dfrac{3\pi}{4}
  4. (4)−3π4-\dfrac{3\pi}{4}

Solution

  1. Multiply by the conjugate: (7+i)(3−4i)32+42\dfrac{(7 + i)(3 - 4i)}{3^{2} + 4^{2}}.
  2. Numerator: 21−28i+3i−4i2=25−25i21 - 28i + 3i - 4i^{2} = 25 - 25i.
  3. So the number is 25−25i25=1−i\dfrac{25 - 25i}{25} = 1 - i.
  4. 1−i1 - i lies in the fourth quadrant with ∣tan⁡θ∣=1|\tan\theta| = 1, so the principal argument is −π4-\tfrac{\pi}{4}.

Answer: −π4-\dfrac{\pi}{4}

Shortcuts and results to remember

  • Any four consecutive powers of ii add up to 0, e.g. i+i2+i3+i4=0i + i^{2} + i^{3} + i^{4} = 0.
  • cis⁡α⋅cis⁡β=cis⁡(α+β)\operatorname{cis}\alpha \cdot \operatorname{cis}\beta = \operatorname{cis}(\alpha + \beta) and cis⁡αcis⁡β=cis⁡(α−β)\dfrac{\operatorname{cis}\alpha}{\operatorname{cis}\beta} = \operatorname{cis}(\alpha - \beta), where cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta = \cos\theta + i\sin\theta.

More practice questions

Practice question 1

The principal argument of −1−i3-1 - i\sqrt{3} is

  1. (1)−2π3-\dfrac{2\pi}{3}
  2. (2)2π3\dfrac{2\pi}{3}
  3. (3)π3\dfrac{\pi}{3}
  4. (4)−π3-\dfrac{\pi}{3}

Solution

  1. x=−1<0x = -1 < 0 and y=−3<0y = -\sqrt{3} < 0, so the point is in the third quadrant
  2. α=tan⁡−1∣yx∣=tan⁡−13=π3\alpha = \tan^{-1}\left|\tfrac{y}{x}\right| = \tan^{-1}\sqrt{3} = \tfrac{\pi}{3}
  3. Third quadrant: arg⁡=−(π−α)=−2π3\arg = -(\pi - \alpha) = -\tfrac{2\pi}{3}

Answer: −2π3-\dfrac{2\pi}{3}

Common mistakes to avoid

  • Taking arg z = tan⁻¹(b/a) without checking the quadrant; for −1 − i this gives π/4 instead of the correct −3π/4.
  • Assuming |z₁ + z₂| = |z₁| + |z₂|; that holds only when z₁ and z₂ have the same argument, otherwise it is only an inequality.
  • Dividing complex numbers by dividing real and imaginary parts separately; multiply by the conjugate of the denominator instead.
  • Giving a principal argument outside (−π, π], such as 7π/4 instead of −π/4.

How to prepare it

For locus questions, substitute z = x + iy and simplify — the curve usually becomes obvious.

Once the formulas feel familiar, test them the way the exam does: practise Complex Numbers questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Complex Numbers chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Complex Numbers part of the EAPCET Mathematics syllabus?
Yes. Complex Numbers is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Complex Numbers for EAPCET?
For locus questions, substitute z = x + iy and simplify — the curve usually becomes obvious.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Complex Numbers?
Taking arg z = tan⁻¹(b/a) without checking the quadrant; for −1 − i this gives π/4 instead of the correct −3π/4. Assuming |z₁ + z₂| = |z₁| + |z₂|; that holds only when z₁ and z₂ have the same argument, otherwise it is only an inequality. Dividing complex numbers by dividing real and imaginary parts separately; multiply by the conjugate of the denominator instead. Giving a principal argument outside (−π, π], such as 7π/4 instead of −π/4.