In this comprehensive tutorial you will learn:

- The nature of a complex number
- Complex Conjugate
- Addition, subtraction, multiplication and division of complex numbers
- Finding complex roots of equations
- Argument and modulus of complex numbers
- Polar form of a complex number
- Locus of points on Argand Diagram
- Euler's relation - complex number in the exponential form
- de Moivre's Theorem
- Trigonometric Identities by de Moivre's Theorem

**E.g.1**

As with everything in maths, complex numbers also have its own birth, may be even in multiple forms! This is one form of it.

Suppose we want to solve the quadratic
equation, **x ^{2}+ x + 6 = 0**. If we
stick to the formula method, the process is as follows:

x = [-1 + √(1 - 24)]/2 or x = [-1 - √(1 - 24)]/2

x = [-1 + √( - 23)]/2 or x = [-1 - √(- 23)]/2

We cannot go beyond this point, as we have come to a stage where we have a square root of a negative number; we have been told it does not exist and we wind up our calculation with familiar conclusions - there are no real roots or solutions. That's it.

However, that was not the end of the world for mathematicians; they have gone much further than that and the concept of complex numbers was born.

In the above example we came across √(-23). Using the rules on indices, we can write it as follows:

√ -23 = √ -1 √ 23

Now we call √-1 as i, and suddenly √-23
becomes something that can be handled.

√-23 = i√ 23

Now, the above quadratic equation can be
solved!

x = (-1 + i√23)/2 or (-1 - i√23)/2

Solutions are expressed in terms of i.
These are called complex numbers.

Complex numbers are expressed in terms of
**i or √-1.**

- i = √-1
- i
^{2}= √-1.√-1 = -1 - i
^{3}= √-1.√-1.√-1 = -i - i
^{4}= √-1.√-1.√-1.√-1 = -1 x -1 = 1

Z = a + ib

a = the real part of the complex number

ib = imaginary part of the complex number

i = √-1

A complex number can be
represented in a x-y grid. Then it is called an **Argand Diagram**: the real part is
shown in the x-axis and the imaginary part in the y-axis.

This is the Cartesian representation of a complex number: the x-axis represents the real part; y-axis represents the imaginary part.

Complex numbers can be added, subtracted, multiplied and even divided like any other number, with a bit of caution though.

**E.g.**

Z_{1} = 8 + i6 ; Z_{2} = 2 - i4

Z_{1} +
Z_{2} = (8 + 2) + i (6 - 4) = 10 + i2

Z_{1} -
Z_{2} = (8 - 2) + i (6 - -4) = 6 + i10

You can practise addition and subtraction interactively with the following applet. Move Z_{1} and Z_{2} so that Z_{3} and Z_{4} represent addition and subtraction respectively.

Z_{1}.Z_{2}

= (8+i6).(2-i4)

= (8X2) - (8X4i) +
(2X6i) - 24i2

= 16 - i32 + i12 -
24(-1) ; note i^{2} = -1

= 16 -i20 +
24

= 40 - i20

If a + bi is a complex number, a - bi is called its complex conjugate.

E.g.

Z = 2 - 3i, so Z^{*} = 2 + 3i

Z = 2 + 3i, so Z^{*} = 2 - 3i

Z = -2 - 3i, so Z^{*} = -2 + 3i

Z_{1} / Z_{2}
= (8 + i6) / (2 - i4)

In this case, we need a special way of dealing with the division;

Multiply both the top and the bottom by the complex conjugate of the bottom. That's it.

Z_{1} / Z_{2}

= (8 + i6) * (2 + i4) / [(2 - i4)(2 + i4)]

= (16 + i32 + i12
+ i224) / (4 + i8 - i8 - i216)

= (-8 + i44) /
(20)

= -2/5 +
i(11/5)

You can practise multiplication and division interactively with the following applet. Move Z_{1} and Z_{2} so that Z_{3} and Z_{4} represent multiplication and division respectively.

Find the square root of 8 + 6i

Let (a + bi)^{2} = 8 + 6i, where a + bi is the square root of 4 + 6i.

So, a^{2} + 2abi - b^{2} = 8 + 6i

Making the real parts on both sides equal,

a^{2} - b^{2} = 8

Making the coefficients of i on both sides equal,

2ab = 6 => b = 6/2a = 3/a

Sub in the other,

a^{2} - 9/a^{2} = 8

a^{4} - 9 = 8a^{2}

a^{4} -8a^{2} -9 = 0

Let y = a^{2}

So, y^{2} -8y -9 = 0

(y - 9)(y + 1)
y = 9 or y = -1 => a^{2} = 9 or a^{2} = -1

Since a^{2} = -1 not valid, a^{2} = 9 => a = ± 3

Sub in b = 3/a => b = 1 or b = -1

So, the square roots of 8 + 6i = (3 + i) or (-3 -i)

The length of the vector that represents the complex number is called its modulus.

If z = a + ib, then modulus is defined as,

√(a2 + b2)

**|Z| =√(a2 + b2)**

The angle between the positive real axis and the vector that represents the complex number is the argument of the complex number.

If arg z = θ, -180 < θ <= 180 or -π < θ <= π

You can practise the concepts with the following animation. Please move the point, Z_{1} with your mouse and see the changes.

Have you noticed how the arg z, changes? It can only take values between -180^{0} and 180^{0} - or its radian equivalents.

If a polynomial equation - quadratic or higher - has complex roots, they are in the form of conjugate pairs - a ± b. Sometimes, one or many be real roots as well - a ±0i.

**E.g.1**

Solve x^{3} - 6x^{2} + 21x - 26 = 0

f(x) = 0

f(2) = 0 => from factor theorem, x-2 is a factor.

Divide x^{3} - 6x^{2} + 21x - 26 by (x-2)

You get, x^{2} - 14x + 13 with no remainder.

Solve, x^{2} - 14x + 13 = 0
x = 2 ± 3i

So, the roots are x =2, x = 2 ± 3i

You can see that there is one real root and two complex roots for the equation - in complex conjugate form.

You can use the following applet to solve polynomials. Just type in the equation in the text box and click the button. Make sure you use, **^**, to represent the powers of x. Use arrow keys, rather than the mouse, while typing in the equations. All roots appear with, **Z**. If they are not visible, just move the grid and look for them!

Complex numbers can be used to find the locus of points on an Argand diagram.

**E.g.1**

The locus of a point, **Z**, which keeps the distance between it and a fixed point, **Z _{1}** , the same is a circle.

In the example, |Z - Z

In order to animate it and see it in action, please press the

|Z - (5 + 4i)| = 3

|Z - 5 - 4i)| = 3

This is a circle with centre, (5,4) and radius = 3.

So, the equation of the locus - the circle in this case - is (x - 5)

**E.g.2**

The locus of a point, **Z**, which keeps the distance between two fixed points, **Z _{1}** and

In the example, the point,

In order to animate it and see it in action, please press the

You can move the point, **Z**, in order to see that it lies on the perpendicular bisector.

Since it is a straight line, the equation of the locus takes y = mx + c form.

Please move the point, **Z _{1}**, to see how the equation of the locus changes. You can use it as an exercise to find the equation of the locus.

Here, the angle t in radians is called the **argument.** Therefore, the complex number can be written in
terms of its modulus and the argument. This is called ** the polar
representation of the complex number**

a = |Z| cos θ : b = |Z| sin t : θ = tan

Z = |Z| cos θ + i|Z| sin θ

If the modulus of a complex number is r and the argument is t, it can be written as (r,θ) in the polar form.

**E.g.1**

|Z| = √(32 + 42) = 5

t = tan

Z = (5,0.92).

**E.g.2**

a = 10 cos 0.3 = 9.5

b= 10 sin 0.3 = 2.9

Z = 9.5 + i2.9

**E.g.3**

tan k = 1/√3

k = π/6

θ = π - k = 5π/6

|Z| = √(3 + 1) = 2

Z = 2(cos 5π/6 + i sin 5π/6)

**E.g.4**

tan k = -1/-1 = 1

k = π/4

θ = π + k or -(π - k);

Since -π< θ <=π

θ = -(π - π/4) = -3π/4 |Z| = √(1 + 1) = √2

Z = √2(cos -3π/4 + i sin -3π/4)

f(x) = f(0) + f'(0)x/1! + f"(0)x^{2}/2! + f"'(0)x^{3}/3! + ....

sin(x) = sin(0) + cos(0)x/1! + -sin(0)x^{2}/2! - cos(0)x^{3}/3! + ...

= x -x^{3}/3! + x^{5}/5!

cos(x) = cos(0) - sin(0)x/1! - cos(0)x^{2}/2! + sin(0)x^{3}/3! + ...

= 1 -x^{2}/3! + x^{4}/4!

e^{x} = 1 + e^{0}x/1! + e^{0}x^{2}/2! + e^{0}x^{3}x/3!+...

= 1 + x + x^{2}/2! + x^{3}/3! + ...

From Maclaurin Series,

e^{iθ} = 1 + iθ/1! + (iθ)^{2}/2! + (iθ)^{3}/3! + ...

= 1 + iθ - θ^{2} -iθ^{3}/3! + θ^{4}/4! + iθ^{5}/5! - θ^{6}/6!

= (1 - θ^{2} + θ^{4}/4! - θ^{6}/6!... ) + i(iθ -iθ^{3}/3! + iθ^{5}/5!...)

= cos θ + i sin θ

e^{iθ} = cos θ + i sin θ

**E.g.1**

Express Z = √3 + i in the form of Z = re^{iθ}

tan θ = 1/√3 => θ = π/6

Z = e

**E.g.2**

Express Z = 3 cos π/3 + i sin π/3 in the form of Z = e^{iθ}.

Z = 3e

**E.g.3**

Express Z = 5 cos π/4 - i sin π/4 in the form of Z = e^{iθ}.

r = 3; and θ=-π/4

Z = 3e

**E.g.4**

Express Z = 4e^{i5π/4} in the form of Z = r[cos θ + i sinθ].

So, Z = 2[cos -π/4 + i sin -π/4]

Let Z_{1} = r_{1}[cosθ_{1}; + i sinθ_{1};] = e^{iθ1;} ;
Z_{2} = r_{2}[cosθ_{2}; + i sinθ_{2};] = e^{iθ2;}

Z_{1} X Z_{2} = r_{1}e^{iθ2;} X r_{2}e^{iθ2;}

= r_{1}r_{2}e^{i[θ1 + θ2]}

= r_{1}r_{2}[cos(θ_{1} + θ_{1}) + i sin(θ_{1} + θ_{2})]

Z_{1} / Z_{2} = r_{1}e^{iθ2;} / r_{2}e^{iθ2;}

= (r_{1}/r_{2}) e^{i[θ1 - θ2]}

= (r_{1}/r_{2}) [cos(θ_{1} - θ_{1}) + i sin(θ_{1} - θ_{2})]

Based on the above proofs, the following set of rules applies in dealing with the product and division of complex numbers:

|Z_{1} Z_{2} | = |Z_{1}||Z_{1}|

arg(Z_{1}Z_{2}) = arg(Z_{1}) + arg(Z_{2})

|Z_{1} / Z_{2} | = |Z_{1}|/|Z_{1}|

arg(Z_{1} / Z_{2}) = arg(Z_{1}) - arg(Z_{2})

**E.g.1**

Express 2(cosπ/6 + i sinπ/6) X 3(cosπ/6 + i sinπ/6)in the form of x + iy.

= 6(cos(π/6+π/6) + i sin(π/6+π/6)

= 6(cosπ/3 + i sinπ/3)

= 3 + 3√3i

**E.g.2**

Express 12(cos5π/6 + i sin5π/6) / 3(cos2π/6 + i sin2π/6)in the form of x + iy.

= 4(cos(5π/6-2π/6) + i sin(5π/6-2π/6)

= 4(cos3π/6 + i sin3π/6)

= 4(cosπ/2 + i sinπ/2)

= 4(0 + i)

= 4i

**E.g.3**

Express 5(cos5π/6 + i sin5π/6) X 2(cos2π/6 - i sin2π/6)in the form of x + iy.

= 5(cos5π/6 + i sin5π/6) X 2(cos(-2π/6) + i sin(-2π/6)) , because, cos-θ = cosθ and sin-θ = -sinθ

= 10(cos3π/6 + i sin3π/6)

= 10(cosπ/2 + i sinπ/2)

= 10(0 + 1i)

= 10i

[r(cosθ + isinθ)]^{n} = r^{n}[cos nθ + i sin nθ]

de Moivre Theorem in exponential form

Z = r[cosθ + isinθ]Z

Z

Z

By multiplying two complex numbers together, it is easy to show the path that leads to **de Moivre Theorem**, which is as follows:

Z

Z

Z

Z

Z

Z

Z

Z

...

...

...

...

Z

This is

de Moivre's Theorem is valid for any **rational** number - negative and positive integers as well as fractions. Normally, it is proved by the **mathematical induction** method.

It is true for any positive integer, n

[r(cos θ + i sinθ)]n=1

Since

Assume that de Moivre's Theorme is true for

Z

Now. let's try it for n = k+1

[r(cos θ + i sinθ)]

= [r

= [r

Therefore, if de Moivre's Theorem is true for n=k, it is true for n=k+1; Since it is true for n=1 too, de Moivre's Theorem is true for any integer, n>=1.

n=0

Since

n is negative =>n<0

Let n=-m, where m is a positive number.Since the theorem is true for positive integers,

= 1 / [r(cos θ + i sinθ)]

= 1 / [r(cos mθ + i msinθ)] (since it is valid for positive integers)

X [cos mθ - i sin mθ] / [cos mθ - i sin mθ] =>

= [cos mθ - i sin mθ] / [(cos mθ - i sin mθ)r

= [cos mθ - i sin mθ] / r

= [cos mθ - i sin mθ] / r

= [cos mθ - i sin mθ] / r

=r

=r

Since -m = n

r

For fractions

Let n=p/q, where p and q are integers.[r(cos θ + i sinθ)]

= r

= r

Now take the q

=r

=r

= r

Therefore, deMoivre's Theorem is valid for fractions as well.

In order to derive the trigonometric identities, we are going to use the **binomial expansion** along with de Moivre Theorem.

(a + x)^{n} = Σ ^{n}C_{r} a^{n-r} x^{r}

**E.g.1**

Find an identity for cos5θ.

[cosθ + i sinθ]^{5} = Σ ^{5}C_{r} cosθ^{5-r} (i sinθ)^{r}

= cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(i sinθ)^{2} + 10 cos^{2}θ(i sinθ)^{3} + 5 cosθ(i sinθ)^{4} + (i sinθ)^{5}

= cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(-sin^{2}θ) + 10 cos^{2}θ(-i sin^{3}θ) + 5 cosθ(sin^{4}θ) + (i sin^{5}θ)

From de Moivre't Theorem,

[cosθ + sinθ]^{5} = [cos5θ + i sin5θ]

So, [cos5θ + i sin5θ] = cos^{5}θ + 5 cos^{4}θ(i sinθ) + 10 cos^{3}θ(-sin^{2}θ) + 10 cos^{2}θ(-i sin^{3}θ) + 5 cosθ(sin^{4}θ) + (i sin^{5}θ)

Now equate the **real** parts on both sides:

cos5θ = cos^{5}θ + 10 cos^{3}θ(-sin^{2}θ) + 5 cosθsin^{4}θ

= cos^{5}θ + 10 cos^{3}θ(-sin^{2}θ) + 5 cosθ(sin^{2}θ)^{2}

= cos^{5}θ -10 cos^{3}θsin^{2}θ + 5 cosθ(1 - cos^{2}θ)^{2}

= cos^{5}θ -10 cos^{3}θ(1 - cos^{2}θ) + 5 cosθ(1 -2cos^{2}θ + cos^{4}θ)

= cos^{5}θ -10 cos^{3}θ + 10 cos^{5}θ + 5 cosθ -10cos^{3}θ +5cos^{5}θ

= 16cos^{5}θ -20 cos^{3}θ + 5 cosθ

**cos 5θ = 16cos ^{5}θ -20 cos^{3}θ + 5 cosθ**

**E.g.2**

Find an identity for sin4θ.

[cosθ + i sinθ]^{4} = Σ ^{4}C_{r} cosθ^{4-r} (i sinθ)^{r}

= cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(i sinθ)^{2} + 4 cosθ(i sinθ)^{3} + (i sinθ)^{4}

= cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(-sin^{2}θ) + 4 cosθ(-i sin^{3}θ) + (i sin^{4}θ)

From de Moivre't Theorem,

[cosθ + sinθ]^{4} = [cos4θ + i sin4θ]

So, [cos4θ + i sin4θ] = = cos^{4}θ + 4 cos^{3}θ(i sinθ) + 6 cos^{2}θ(-sin^{2}θ) + 4 cosθ(-i sin^{3}θ) + (i sin^{4}θ)

Now equate the **imaginary** parts on both sides:

sin4θ = 4cos^{3}θsinθ - 4 cosθ sin^{3}θ + sin^{4}θ

**sin 4θ = 4cos ^{3}θsinθ - 4 cosθ sin^{3}θ + sin^{4}θ**

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7^{th} edition in print.