An Interactive Introduction to Complex Numbers

Jens Siebel (University of Applied Sciences Kaiserslautern) An Interactive Introduction to Complex Numbers

1. Introduction We know that some polynomial equations do not have any solutions on R/ .

Example 1.1: Solve x2 +1 = 0 for x : x2+=10 ⇔ x 2 =− 1 ⇔ xx =−−∨=− 11

Problem: The equation has no solution on R/ , as −1 ∉R / .

An exit strategy is to introduce an “imaginary unit“ i with i = − 1 or i2 = − 1, alternatively. Then, x2 +1 = 0 has two solutions: x= − i and x= i .

Rule 1.1 The imaginary unit i is defined as i = − 1 or i2 = − 1.

Example 1.2: Solve x2 −2 ⋅ x + 5 = 0 for x :

2 2 2 −−22  −− 22 xx−⋅+=⇔=−250 x − −∨=− 5 x  + −⇔ 5 22  22 xx=−−∨=+−14 14 ⇔ x =− 141 ⋅−∨=+() x 141 ⋅−() ⇔ x=−⋅−∨=+⋅−121 x 121 ⇔ xixi =−⋅∨=+⋅ 12 12

The solutions x=1 − 2 ⋅ i and x=1 + 2 ⋅ i are “complex“ numbers. In cartesian form they are composed from a real number and an imaginary number. Normally, complex numbers are denoted z . According to the fundamental theorem of algebra, any polynomial equation of grade n∈ N / has n complex solutions.

1 Jens Siebel: An Interactive Introduction to Complex Numbers 1. Introduction Cartesian form Rule 1.2 A complex number z in cartesian form is written z= a + bi ⋅ , a, b∈ R / with • a= Re( z ) : real part of z • b= Im( z ) : imaginary part of z

• i : imaginary unit.

Example 1.3:

z1 =1 − 2 ⋅ i and z2 =1 + 2 ⋅ i are the solutions of example 1.2. We have

a1=Re( z 1 ) = 1 , b1=Im( z 1 ) = − 2

a2=Re( z 2 ) = 1 , b2=Im( z 2 ) = 2

Rule 1.3 C/ is the set of complex numbers z . It comprises all other sets of numbers.

N/ Z/ Q/ R/ C/

Figure 1.1: Sets of numbers

Any real number x is a complex number z= x +0 ⋅ i . Real numbers do not have imaginary units.

Rule 1.4 Complex numbers are presented as points or position vectors in a cartesian diagram (“complex plane”). Any real number is located on the real axis.

2 Jens Siebel: An Interactive Introduction to Complex Numbers 1. Introduction

Im(z)

z= a + bi ⋅ b⋅ i

Re(z) a Figure 1.2: Complex plane

Example 1.4:

Open the Applet Basic Calculations and enter z1 =1 + 2 ⋅ i and z2 =−0.5 − 1.5 ⋅ i in cartesian form.

Polar form In studying the Applet Basic Calculations in example 1.4 you may have noticed that the complex number z1 =1 + 2 ⋅ i has also been displayed as

63.4349 °i z1 =2.2361 ⋅ exp( 63.4349 °⋅= i) 2.2361 ⋅ e . This is the so-called “exponential form” - a special case of the “polar form”. Vector algebra allows us to turn any complex number z from cartesian form into polar form.

3 Jens Siebel: An Interactive Introduction to Complex Numbers 1. Introduction

Im(z)

b⋅ i z= a + bi ⋅

a Re(z) Figure 1.3: Geometric presentation of z in polar form

In order to rewrite z= a + bi ⋅ in polar form we have to determine the value of ϕ and the length r of the position vector z= a + bi ⋅ .

Rule 1.5 z ≥ 0 is the absolute value of a complex number. It corresponds the length r∈ R / ≥0 of vector z= a + bi ⋅ . Applying Pythagoras’ equation we get zr= = a2 + b 2 .

Rule 1.6 ϕ is the angle between the positive real axis and the position vector z= a + bi ⋅ . ϕ is also called the „argument of z “: ϕ = arg z . It can be given both in degree or radian.

Due to vector algebra we have a= r ⋅ cos ϕ and b= r ⋅ sin ϕ . Then, we can rewrite z= a + bi ⋅ as zr=⋅cosϕ +⋅⋅ ri sin ϕ =⋅ r( cos ϕϕ +⋅ i sin ) .

Rule 1.7 zr=⋅cosϕ +⋅⋅ ri sin ϕ =⋅ r( cos ϕϕ +⋅ i sin ) is the “trigonometric form“ of complex number z= a + bi ⋅ with a= r ⋅ cos ϕ and b= r ⋅ sin ϕ .

b Regarding figure 1.3 we see that tan ϕ = and hence a

4 Jens Siebel: An Interactive Introduction to Complex Numbers 1. Introduction  b arctanifa> 0, b ≥ 0ϕ ∈°°[ 0 , 90 )  a  90°ifa = 0, b > 0  b ϕ =arg z = 180°+ arctanif a < 0ϕ ∈°°() 90 , 270 (degree)  a 270°ifa = 0, b < 0  b 360°+ arctanifa >≤ 0, b 0ϕ ∈°°( 270 , 360 ]  a or

 b π  arctanifa> 0, b ≥ 0ϕ ∈ 0,   a  2     π ifa=0, b > 0  2   b π 3  ϕπ==+argz arctan ifa <∈⋅ 0 ϕπ ,  (radian).  a 2 2   3  ⋅π ifa =0, b < 0  2  b 3  2⋅+π arctanifa >≤ 0, b 0 ϕππ ∈⋅⋅ , 2   a 2 

Example 1.5:

The trigonometric form of z1 =1 + 2 ⋅ i in example 1.4 is 2 z=2.2361 ⋅( cos63.4349 °+⋅ i sin 63.4349 ° ) , as ϕ =arctan = 63.4349 ° and 1 1 1

2 2 r1 =1 + 2 = 2.2361 .

The Euler equation enables us to convert z=⋅ r(cosϕ +⋅ i sin ϕ ) into z= r ⋅ e ϕ⋅i .

Rule 1.8 z= r ⋅ e ϕ⋅i is the exponential form of complex number z .

In General The polar form of z is depicted by r and ϕ . It can be displayed in either trigonometric form or exponential form. Note that it requires the trigonometric form to turn the exponential form into the cartesian form.

5 Jens Siebel: An Interactive Introduction to Complex Numbers 1. Introduction Throughout the following and in the applets we will prefer the exponential form to the trigonometric form, as the first one has a shorter notation and a less demanding algebra. Note that both applets display the exponential form as z= r ⋅exp (ϕ ⋅ i ) .

Example 1.6:

45 °i 2.5 ⋅i Open the Applet Basic Calculations and enter z1 =1 ⋅ e and z2 =2 ⋅ e in exponential form. Note that ϕ1=argz 1 = 45 ° is in degree, whereas ϕ2=argz 2 = 2.5 is in radian. When entering ϕ in degree you have to add “°”.

Example 1.7:

45 °i 2.5 ⋅i In order to rewrite z1 =1 ⋅ e and z2 =2 ⋅ e from example 1.6 in cartesian form, one has to calculate as follows:

45 °i ze1 =⋅=⋅1 1( cos 45 °+⋅ i sin 45 °=⋅) 1( 0.7071 +⋅ i 0.7071) = 0.7071 + 0.7071 ⋅ i

2.5 ⋅i ze2 =⋅=⋅2 2( cos 2.5 +⋅ i sin 2.5) =− 1.6023 + 1.1969 ⋅ i

Exercise 1.1: Rewrite z=+⋅13, iz =−+⋅ 13, iz =−−⋅ 13 i and z=1 − 3 ⋅ i in exponential form. Use the Applet Basic Calculations to check your answers.

Exercise 1.2: Rewrite z=⋅2 e75°i , zez = 90 ° i ,2 =⋅ e 285 ° i and z=2 ⋅ e 0.25 ⋅πi in cartesian form. Use the

Applet Basic Calculations to check your answers. You can enter 2 as “sqrt(2)“ and 0.25 ⋅π as “0.25*pi“ or “pi/4”.