Complex Numbers Notes FE Sem 1 Applied Maths-I | Mumbai Univ

This module revises the algebra of complex numbers and their Cartesian, polar and exponential forms, then puts De Moivre's theorem to work. You will learn to expand cos nθ and sin nθ in powers of cos θ and sin θ, to write powers like cosⁿθ and sinⁿθ as sums of cosines or sines of multiple angles, and to find all powers and roots of a complex number. These skills are used later for AC circuit phasors, signal analysis and for the hyperbolic and logarithmic functions of Module 2.

Summary

Pre-requisite review: a complex number is \( z = x + iy \) with \( i^2 = -1 \); x is the real part and y the imaginary part. Addition, subtraction, multiplication and division follow ordinary algebra with \( i^2 = -1 \); division is done by multiplying numerator and denominator by the conjugate \( \bar z = x - iy \). The modulus is \( |z| = r = \sqrt{x^2 + y^2} \) and the argument is \( \theta \) with \( \tan\theta = y/x \), chosen in the correct quadrant.

Forms of a complex number: Cartesian form \( x + iy \), polar form \( r(\cos\theta + i\sin\theta) \) and exponential form \( re^{i\theta} \), linked by Euler's formula \( e^{i\theta} = \cos\theta + i\sin\theta \). In polar or exponential form multiplication multiplies moduli and adds arguments, and division divides moduli and subtracts arguments.

De Moivre's theorem (statement): for any rational n, \( (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \). For integer n this value is unique; for fractional n = p/q it is one of the q values of the power.

Expansion of cos nθ and sin nθ in powers of cos θ and sin θ (1.1): write \( \cos n\theta + i\sin n\theta = (c + is)^n \) with \( c = \cos\theta, s = \sin\theta \), expand by the binomial theorem and equate real and imaginary parts. For example \( \cos 3\theta = 4c^3 - 3c \) and \( \sin 3\theta = 3s - 4s^3 \). Replacing \( s^2 = 1 - c^2 \) or \( c^2 = 1 - s^2 \) gives the result purely in cos θ or sin θ.

Expansion of cosⁿθ and sinⁿθ in multiples of θ (1.1): put \( x = \cos\theta + i\sin\theta \), so \( x + 1/x = 2\cos\theta \), \( x - 1/x = 2i\sin\theta \), \( x^m + 1/x^m = 2\cos m\theta \) and \( x^m - 1/x^m = 2i\sin m\theta \). Expand \( (x + 1/x)^n \), \( (x - 1/x)^n \) or their product, group terms in pairs \( x^m \pm x^{-m} \) and convert back. This turns a power such as \( \cos^5\theta \) into a sum of cosines of multiple angles, which is useful in integration and Fourier analysis.

Powers of a complex number (1.2): convert to polar form, \( z = r(\cos\theta + i\sin\theta) \), then \( z^n = r^n(\cos n\theta + i\sin n\theta) \). This is much faster than repeated binomial multiplication, e.g. \( (1 + i)^8 = (\sqrt2)^8(\cos 2\pi + i\sin 2\pi) = 16 \).

Roots of a complex number (1.2): since the argument is unchanged by adding \( 2k\pi \), \( z^{1/n} = r^{1/n}\left[\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n}\right] \), k = 0, 1, ..., n − 1. There are exactly n distinct nth roots; they lie on a circle of radius \( r^{1/n} \) and are equally spaced at angles \( 2\pi/n \). The same idea solves equations such as \( x^n = a \) and polynomial equations that can be reduced to that form.

Self-learning topic: basics of complex numbers (geometric representation in the Argand plane, conjugates, modulus and argument rules). These are assumed throughout the module and in Module 2.

Mind map

Method: expressing cosⁿθ or sinⁿθ in multiples of θ

Important definitions

Formulas

Key points to remember

• Always find the argument in the correct quadrant; tan⁻¹(y/x) alone gives the wrong angle for points in the 2nd and 3rd quadrants.
• For cos nθ, sin nθ in powers use (c + is)^n; for cosⁿθ, sinⁿθ in multiples use x = cos θ + i sin θ.
• Powers of i cycle: i, −1, −i, 1; (2i)^n gives a factor iⁿ that must be handled carefully.
• Any power of cos θ, and any even power of sin θ, expands into cosines of multiple angles only; an odd power of sin θ expands into sines only.
• An nth root problem always has exactly n distinct answers, k = 0 to n − 1.
• The n roots of z have modulus r^{1/n} and are spaced 2π/n apart; the n roots of unity add up to 0.
• To solve an equation like x⁴ − x³ + x² − x + 1 = 0, multiply by (x + 1) to get x⁵ + 1 = 0 and then reject the extra root x = −1.
• Check every expansion by substituting θ = 0, π/2 or a small θ.

Solved examples

Example 1. Express \( \cos^5\theta \) in terms of cosines of multiples of θ.

Let \( x = \cos\theta + i\sin\theta \). Then \( x + \frac1x = 2\cos\theta \) and \( x^m + \frac{1}{x^m} = 2\cos m\theta \).

\( 2^5\cos^5\theta = \left(x + \frac1x\right)^5 = x^5 + 5x^3 + 10x + \frac{10}{x} + \frac{5}{x^3} + \frac{1}{x^5} \).

Grouping: \( = \left(x^5 + \frac{1}{x^5}\right) + 5\left(x^3 + \frac{1}{x^3}\right) + 10\left(x + \frac1x\right) = 2\cos5\theta + 10\cos3\theta + 20\cos\theta \).

Hence \( \cos^5\theta = \frac{1}{16}(\cos5\theta + 5\cos3\theta + 10\cos\theta) \).

Check θ = 0: (1 + 5 + 10)/16 = 1 = cos⁵0.

Example 2. Prove that \( \cos^4\theta\sin^3\theta = -\frac{1}{64}(\sin7\theta + \sin5\theta - 3\sin3\theta - 3\sin\theta) \).

Let \( x = \cos\theta + i\sin\theta \); then \( 2\cos\theta = x + \frac1x \), \( 2i\sin\theta = x - \frac1x \).

\( (2\cos\theta)^4(2i\sin\theta)^3 = 2^7 i^3\cos^4\theta\sin^3\theta = -128\,i\cos^4\theta\sin^3\theta \).

RHS of the product: \( \left(x + \frac1x\right)^4\left(x - \frac1x\right)^3 = \left(x + \frac1x\right)\left(x^2 - \frac{1}{x^2}\right)^3 \).

\( \left(x^2 - \frac{1}{x^2}\right)^3 = x^6 - 3x^2 + \frac{3}{x^2} - \frac{1}{x^6} \).

Multiplying by \( x + \frac1x \): \( x^7 + x^5 - 3x^3 - 3x + \frac3x + \frac{3}{x^3} - \frac{1}{x^5} - \frac{1}{x^7} \) \( = \left(x^7 - \frac{1}{x^7}\right) + \left(x^5 - \frac{1}{x^5}\right) - 3\left(x^3 - \frac{1}{x^3}\right) - 3\left(x - \frac1x\right) \).

\( = 2i(\sin7\theta + \sin5\theta - 3\sin3\theta - 3\sin\theta) \).

So \( -128\,i\cos^4\theta\sin^3\theta = 2i(\sin7\theta + \sin5\theta - 3\sin3\theta - 3\sin\theta) \), giving the required result after dividing by −128i.

Example 3. Prove that \( \sin5\theta = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta \).

By De Moivre's theorem, \( \cos5\theta + i\sin5\theta = (c + is)^5 \) where \( c = \cos\theta, s = \sin\theta \).

Binomial expansion: \( (c + is)^5 = c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5 \).

Equating imaginary parts: \( \sin5\theta = 5c^4s - 10c^2s^3 + s^5 \).

Put \( c^2 = 1 - s^2 \): \( \sin5\theta = 5(1 - s^2)^2s - 10(1 - s^2)s^3 + s^5 \).

\( = 5s - 10s^3 + 5s^5 - 10s^3 + 10s^5 + s^5 = 5s - 20s^3 + 16s^5 \).

Hence \( \sin5\theta = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta \).

Example 4. If α and β are the roots of \( x^2 - 2x + 4 = 0 \), prove that \( \alpha^n + \beta^n = 2^{n+1}\cos\frac{n\pi}{3} \).

Roots: \( x = \frac{2 \pm \sqrt{4 - 16}}{2} = 1 \pm i\sqrt3 \).

Polar form: modulus \( \sqrt{1 + 3} = 2 \), argument \( \tan^{-1}\sqrt3 = \pi/3 \). So \( \alpha = 2\left(\cos\frac\pi3 + i\sin\frac\pi3\right) \), \( \beta = 2\left(\cos\frac\pi3 - i\sin\frac\pi3\right) \).

By De Moivre: \( \alpha^n = 2^n\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right) \), \( \beta^n = 2^n\left(\cos\frac{n\pi}{3} - i\sin\frac{n\pi}{3}\right) \).

Adding: \( \alpha^n + \beta^n = 2\cdot2^n\cos\frac{n\pi}{3} = 2^{n+1}\cos\frac{n\pi}{3} \).

Example 5. Find the cube roots of \( -8i \).

\( -8i = 8\left[\cos\left(-\frac\pi2\right) + i\sin\left(-\frac\pi2\right)\right] = 8\left[\cos\left(2k\pi - \frac\pi2\right) + i\sin\left(2k\pi - \frac\pi2\right)\right] \).

\( (-8i)^{1/3} = 2\left[\cos\frac{(4k - 1)\pi}{6} + i\sin\frac{(4k - 1)\pi}{6}\right] \), k = 0, 1, 2.

k = 0: \( 2\left(\cos\frac\pi6 - i\sin\frac\pi6\right) = \sqrt3 - i \).

k = 1: \( 2\left(\cos\frac\pi2 + i\sin\frac\pi2\right) = 2i \).

k = 2: \( 2\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right) = -\sqrt3 - i \).

Check: \( (2i)^3 = 8i^3 = -8i \).

Example 6. Solve \( x^4 - x^3 + x^2 - x + 1 = 0 \).

Multiply by (x + 1): \( (x + 1)(x^4 - x^3 + x^2 - x + 1) = x^5 + 1 = 0 \), so \( x^5 = -1 = \cos(2k + 1)\pi + i\sin(2k + 1)\pi \).

\( x = \cos\frac{(2k + 1)\pi}{5} + i\sin\frac{(2k + 1)\pi}{5} \), k = 0, 1, 2, 3, 4.

k = 2 gives \( x = \cos\pi = -1 \), which is the extra root introduced by the factor (x + 1), so it is rejected.

Required roots (k = 0, 1, 3, 4): \( \cos\frac\pi5 \pm i\sin\frac\pi5 \) and \( \cos\frac{3\pi}{5} \pm i\sin\frac{3\pi}{5} \) (k = 4 and k = 3 give the conjugates of k = 0 and k = 1).

Example 7. Find the continued product of all four values of \( \left(\frac12 + i\frac{\sqrt3}{2}\right)^{3/4} \).

\( \frac12 + i\frac{\sqrt3}{2} = \cos\frac\pi3 + i\sin\frac\pi3 \); cubing, \( \left(\cos\frac\pi3 + i\sin\frac\pi3\right)^3 = \cos\pi + i\sin\pi \).

So the required values are \( (\cos\pi + i\sin\pi)^{1/4} = \cos\frac{(2k + 1)\pi}{4} + i\sin\frac{(2k + 1)\pi}{4} \), k = 0, 1, 2, 3.

Product = \( \cos\frac{(1 + 3 + 5 + 7)\pi}{4} + i\sin\frac{(1 + 3 + 5 + 7)\pi}{4} = \cos4\pi + i\sin4\pi = 1 \).

Hence the continued product is 1.

Important questions and answers

Q1. State De Moivre's theorem and use it to find \( (1 + i)^8 \). (3 marks)

Statement: for rational n, \( (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \) (for fractional n this is one of the values).

\( 1 + i = \sqrt2\left(\cos\frac\pi4 + i\sin\frac\pi4\right) \).

\( (1 + i)^8 = (\sqrt2)^8(\cos2\pi + i\sin2\pi) = 16(1 + 0) = 16 \).

Q2. Expand \( \cos6\theta \) in powers of \( \cos\theta \). (4 marks)

\( \cos6\theta = \text{Re}(c + is)^6 = c^6 - 15c^4s^2 + 15c^2s^4 - s^6 \), with \( c = \cos\theta, s = \sin\theta \).

Put \( s^2 = 1 - c^2 \): \( = c^6 - 15c^4(1 - c^2) + 15c^2(1 - c^2)^2 - (1 - c^2)^3 \).

\( = c^6 - 15c^4 + 15c^6 + 15c^2 - 30c^4 + 15c^6 - 1 + 3c^2 - 3c^4 + c^6 \).

\( \cos6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 \). Check θ = 0: 32 − 48 + 18 − 1 = 1.

Q3. Prove that \( \sin^4\theta\cos^2\theta = \frac{1}{32}(\cos6\theta - 2\cos4\theta - \cos2\theta + 2) \). (5 marks)

Let \( x = \cos\theta + i\sin\theta \). Then \( (2i\sin\theta)^4(2\cos\theta)^2 = 64\sin^4\theta\cos^2\theta \) since \( i^4 = 1 \).

\( \left(x - \frac1x\right)^4\left(x + \frac1x\right)^2 = \left(x^2 - \frac{1}{x^2}\right)^2\left(x - \frac1x\right)^2 = \left(x^4 - 2 + \frac{1}{x^4}\right)\left(x^2 - 2 + \frac{1}{x^2}\right) \).

Expanding: \( x^6 - 2x^4 - x^2 + 4 - \frac{1}{x^2} - \frac{2}{x^4} + \frac{1}{x^6} = \left(x^6 + \frac{1}{x^6}\right) - 2\left(x^4 + \frac{1}{x^4}\right) - \left(x^2 + \frac{1}{x^2}\right) + 4 \).

\( = 2\cos6\theta - 4\cos4\theta - 2\cos2\theta + 4 \).

Dividing by 64 gives the result. Check θ = π/4: LHS = 1/8, RHS = (0 + 2 − 0 + 2)/32 = 1/8.

Q4. Prove that \( \frac{\sin6\theta}{\sin\theta} = 32\cos^5\theta - 32\cos^3\theta + 6\cos\theta \). (4 marks)

\( \sin6\theta = \text{Im}(c + is)^6 = 6c^5s - 20c^3s^3 + 6cs^5 \).

Divide by s: \( \frac{\sin6\theta}{\sin\theta} = 6c^5 - 20c^3s^2 + 6cs^4 \).

Put \( s^2 = 1 - c^2 \): \( 6c^5 - 20c^3(1 - c^2) + 6c(1 - 2c^2 + c^4) = 6c^5 - 20c^3 + 20c^5 + 6c - 12c^3 + 6c^5 \).

\( = 32c^5 - 32c^3 + 6c \).

Q5. If \( x + \frac1x = 2\cos\theta \), prove that \( x^n + \frac{1}{x^n} = 2\cos n\theta \). (3 marks)

\( x^2 - 2x\cos\theta + 1 = 0 \Rightarrow x = \cos\theta \pm i\sin\theta \).

Take \( x = \cos\theta + i\sin\theta \); then \( 1/x = \cos\theta - i\sin\theta \) (the other sign just interchanges x and 1/x).

By De Moivre: \( x^n = \cos n\theta + i\sin n\theta \), \( x^{-n} = \cos n\theta - i\sin n\theta \).

Adding: \( x^n + x^{-n} = 2\cos n\theta \).

Q6. Find all the values of \( (1 + i)^{1/3} \). (4 marks)

\( 1 + i = \sqrt2\left[\cos\left(2k\pi + \frac\pi4\right) + i\sin\left(2k\pi + \frac\pi4\right)\right] \).

\( (1 + i)^{1/3} = 2^{1/6}\left[\cos\frac{(8k + 1)\pi}{12} + i\sin\frac{(8k + 1)\pi}{12}\right] \), k = 0, 1, 2.

Values: \( 2^{1/6}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right) \), \( 2^{1/6}\left(\cos\frac{9\pi}{12} + i\sin\frac{9\pi}{12}\right) = 2^{1/6}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) \), \( 2^{1/6}\left(\cos\frac{17\pi}{12} + i\sin\frac{17\pi}{12}\right) \).

Q7. Solve \( x^6 + 1 = 0 \). (4 marks)

\( x^6 = -1 = \cos(2k + 1)\pi + i\sin(2k + 1)\pi \).

\( x = \cos\frac{(2k + 1)\pi}{6} + i\sin\frac{(2k + 1)\pi}{6} \), k = 0 to 5.

k = 0, 5: \( \cos\frac\pi6 \pm i\sin\frac\pi6 = \frac{\sqrt3}{2} \pm \frac i2 \).

k = 1, 4: \( \cos\frac\pi2 \pm i\sin\frac\pi2 = \pm i \).

k = 2, 3: \( \cos\frac{5\pi}{6} \pm i\sin\frac{5\pi}{6} = -\frac{\sqrt3}{2} \pm \frac i2 \).

Q8. Prove that \( (1 + i\sqrt3)^n + (1 - i\sqrt3)^n = 2^{n+1}\cos\frac{n\pi}{3} \) and hence find the value for n = 6. (4 marks)

\( 1 \pm i\sqrt3 = 2\left(\cos\frac\pi3 \pm i\sin\frac\pi3\right) \).

By De Moivre the sum equals \( 2^n\cdot2\cos\frac{n\pi}{3} = 2^{n+1}\cos\frac{n\pi}{3} \).

For n = 6: \( 2^7\cos2\pi = 128 \).

Q9. Express \( \sin^7\theta \) in terms of sines of multiples of θ. (6 marks)

Let \( x = \cos\theta + i\sin\theta \). \( (2i\sin\theta)^7 = 2^7i^7\sin^7\theta = -128\,i\sin^7\theta \) since \( i^7 = -i \).

\( \left(x - \frac1x\right)^7 = x^7 - 7x^5 + 21x^3 - 35x + \frac{35}{x} - \frac{21}{x^3} + \frac{7}{x^5} - \frac{1}{x^7} \).

\( = \left(x^7 - \frac{1}{x^7}\right) - 7\left(x^5 - \frac{1}{x^5}\right) + 21\left(x^3 - \frac{1}{x^3}\right) - 35\left(x - \frac1x\right) \).

\( = 2i(\sin7\theta - 7\sin5\theta + 21\sin3\theta - 35\sin\theta) \).

So \( \sin^7\theta = -\frac{1}{64}(\sin7\theta - 7\sin5\theta + 21\sin3\theta - 35\sin\theta) = \frac{1}{64}(35\sin\theta - 21\sin3\theta + 7\sin5\theta - \sin7\theta) \).

Q10. Show that the sum of the n nth roots of unity is zero. (3 marks)

The nth roots of unity are \( 1, \omega, \omega^2, \dots, \omega^{n-1} \) where \( \omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n} \neq 1 \) (n > 1).

Sum = \( \frac{1 - \omega^n}{1 - \omega} \) (geometric series).

Since \( \omega^n = \cos2\pi + i\sin2\pi = 1 \), the sum is 0.

Exam tips

• Begin every expansion answer with 'Let x = cos θ + i sin θ' or 'By De Moivre's theorem'; MU examiners award a mark for the correct setup.
• Keep track of the power of i on the left side (iⁿ cycles every 4); most sign errors in sinⁿθ problems come from here.
• Write the general argument θ + 2kπ before dividing by n and list all values of k; missing roots loses marks.
• For polynomial equations, state clearly which root is rejected and why.
• Verify your final expansion by substituting θ = 0 or θ = π/2; it takes 20 seconds and catches coefficient mistakes.
• Leave answers in exact form (cos π/12 etc.) unless decimals are asked for.

FAQs

When do I use (c + is)^n and when x = cos θ + i sin θ?

Use (c + is)^n with the binomial theorem to express cos nθ or sin nθ as powers of cos θ and sin θ. Use x = cos θ + i sin θ with x ± 1/x to express powers such as cos⁵θ or sin⁴θ cos²θ as sums of multiple angles.

How many values does z^(p/q) have?

If p/q is in lowest terms, z^(p/q) has exactly q distinct values, obtained by taking k = 0, 1, ..., q − 1 in the general argument.

Is De Moivre's theorem proved in the syllabus?

No. The MU NEP syllabus lists only the statement of De Moivre's theorem as a prerequisite; questions focus on its applications to expansions, powers and roots.

Why must the argument be taken in the correct quadrant?

tan⁻¹(y/x) gives the same value for (x, y) and (−x, −y). Using it blindly for points in the 2nd or 3rd quadrant gives roots that are wrong by π, so always check the signs of x and y.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Mathematics-I) and were prepared with AI help and checked.