Argand Plane and Modulus Argument Form

Let's start with some terminology

If z = x +iy

Re(z) = x (real component of z)

Im(z) = y (imaginary component of z)

The Argand plane (also known as the complex plane) is used to plot complex numbers. Complex numbers 'z' are plotted on the Argand plane with Re(z) on the x axis and Im (z) on the y axis.

Eg.

A represents 2+i, B represents 1, C represents -1 +3i, D represents -1-2i.

Now let's look at absolute values again.

The absolute value of a number, also called the modulus of a number is the distance of the number from the origin.

If the number is a real number, then the distance from the origin is just the number without the sign.

Eg. 1

|4|= 4,

Eg. 2

|-3|=3

For a complex number, the modulus can be found using the distance formula.

Eg. 1

|3+4i| = √(3² +4²) = 5

 

Eg. 2

|5-12i|= √(5² + 12²) = 13

By definition, if z = x +iy, |x +iy|= √(x² +y²) , this is known as the modulus of z or sometimes shortened to mod z.

It is not enough to define a complex number by its modulus because there are infinite complex numbers that have the same modulus.

So we also define complex numbers by the angle that is created with the x-axis when that point is joined to the origin (known as the argument, often written as arg).

We can find the argument by using the fact that m(gradient) = tan(θ)

Eg. 1

If z = 1 + i√3

First we should determine which quadrant z is in. In this case z is in the first quadrant.

tanθ = √3/1

θ = π/3 + 2πn or 4π/3 +2πn (where n is an integer)

As we can see, there are infinite solutions for θ.

We know that 1 + i√3 is in the first quadrant and therefore 4π/3 + 2πn will not be solutions for θ, but we still have infinite possible solutions, so we limit the range of the answers to –π ≤ θ ≤ π.

So the final answer is arg z = π/3, this is known as the principle arguments, and the other solutions for θ are known as arguments.

Eg. 2

If z = –1 + i 

 

z is in the second quadrant

tanθ = 1/-1

θ = 3π/4 or -π/4 (for –π ≤ θ ≤ π)

Since z is in the second quadrant, arg z ≠ 7π/4 +2πn

Therefore arg z = 3π/4

Eg. 3

If z = -3 – 3i

z is in the third quadrant

tanθ = -3/-3

θ = π/4 or -3π/4 (for –π ≤ θ ≤ π)

Since z is in the third quadrant, arg z = -3π/4.

Now that we know how to work out the modulus and argument of a complex number, we can write it in modulus argument form.

The point z represents that complex number z = x + iy

Let θ = arg z and |z|= r

Therefore x = r cosθ and y = r sinθ

Therefore x +iy = r (cosθ + isinθ), sometimes written as r cisθ, (cis = cos + isin)

This is known as the modulus argument form. The reason why we write complex numbers in this form is that there are various properties that allow use to easily perform operations in complex numbers (will be shown in the next post).

The modulus and argument are sometimes known as polar coordinates. They are written in the form [r, θ] where r is the modulus and θ is the argument. Please note that square brackets are used instead of parentheses for polar coordinates.

Square Roots of Complex Numbers

Complex numbers can also be square rooted.

Example 1

Find the square root of 3 +4i

(a+ib)² = 3 +4i

a² + 2iab - b² = 3 +4i

By equating real and imaginary parts

a² - b² = 3

2ab = 4

By solving simultaneously

a = 2 or -2

When a = 2, b =1

When a = -2, b = -1

So 2 + i and -2 - i are the square roots of 3 +4i

Example 2

Find the square root of i

(a +ib)² = i

a² +2iab -b² = i

By equating real and imaginary parts

a² -b² = 0 (1)

2ab =1

a = 1/2b (2)

Substitute (2) into (1)

∴ (1/2b)² - b² = 0

1/4b² - b² = 0

1 - 4b⁴ = 0

(1+2b²)(1-2b²) = 0

(1+2b²)(1-√2b)(1+√2b)=0

b = 1/√2 or -1/√2

When b = 1/√2, a = 1/√2

When b = -1/√2, a = -1/√2

Square roots of i are ±(1/√2 + i/√2)

With this, we are able to solve any quadratic equation even if the coefficients are complex.

Example

x² -(1+2i)x - (1 - i) =0

Δ = b² - 4ac

= (1+2i)² -4(-1+2i)

= 1+4i-4 +4 -4i

= 1 (in this case Δ is a real number, there are cases where Δ is complex and the above method would have to be used to find the roots to Δ)

x = (-b ± √Δ)/2a

= [(1+2i) ± 1]/2

= i or 1+i

Introduction to Complex Numbers

Suppose we wanted to take the square √ of -1. We know that there aren't any real numbers that would give a suitable answer so mathematicians introduced a number "i" known as the imaginary unit.

i² = -1

√-1 = i

√-4 = 2i

√ -3 = i√3

These numbers are known as imaginary numbers. They are written in the form ib (where b is a real number. b is placed in front of i if it is a rational number, otherwise it is placed on the back)

A complex number is a number that has both a real and a imaginary component. They are written in the form a+ ib (where a and b are real numbers)

All numbers are complex numbers. Real numbers are simply complex numbers with a zero imaginary component. Eg 1 = 1+ 0i. Imaginary numbers are complex numbers with a zero real component. Eg. 4i = 0 +4i

Equal complex numbers

Two complex numbers are equal only if their real components are equal and their imaginary components are also equal.

Eg. (a + ib) = (c + id) iff a = c and b = d.

Fyi. Iff is a mathematical term which means if and only if.

An example question can be found here.


Quadratic with negative discriminant

With complex numbers we can solve any quadratic equation even if the discriminant is less than 0.

Eg.  x² + x + 1 =0

x = (-1 ± √-3)/2

x = -1/2 + i√3/2 or 1/2 - i√3/2

Conjugate of a complex number

If z = a + ib the conjugate of z, written as

= a – ib

z= (a+ ib) (a –ib) = a² – i²b²

=a² + b²

The conjugate of a complex number is used to change the denominator into a real number if it is a complex number.

Eg. 2/1-2i = 2 (1+2i)/(1-2i)(1+2i) = (2 + 4i)/5

Sum, difference, product and division of complex numbers.

You do not have to remember these as they can be easily worked out with actual numbers.

Sum:

(a+ib) + (c+id) = (a+c) + i(b+d)

Difference:

(a+ib) – (c+id) = (a-c) + i(b-d)

Product:

(a+ib)(c+id) = ac + iad + ibc + i²bd

=ac – bd + i(ad+bc)

Division:

(a+ib)/(c+id) = (a+ib)(c-id)/(c+id)(c-id) (multiply by conjugate)

= (ac –iad +ibc –i²bd)/(c² +d²)

= [ac + bd + i(bc – ad)]/(c² +d²)

Example questions

Let z = 5 –i, find:

i)           z² in the form of a+ib

ii)          z + 2 in the form of a+ib

iii)         i/z in the form of a+ib

Solution

i)      z² = (5-i)²

= 25 – 10i + i²

= 24 – 10i

ii)      = 5 +i

z + 2 = (5-i) + 2(5+i)

= 5 – i + 10 +2i

= 15 + i

iii)     i/z = i/(5-i)

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

= (-1 + 5i)/26

= -1/26 + 5/26 i