Enter your Sign on user name and password.

Forgot password?
Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Physical Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Sat Jun 27, 2015 5:58 PM

Post by Jinbin Chen on June 27, 2015

Hi, Mr. Raffi! Is differential equation a necessary prerequisite before viewing the quantum mechanics lectures in this series? I have had some exposure to multivariable calculus and linear algebra from your lecture series on this website (and I appreciate your enthusiasm in those videos, especially in linear algebra), but I am not sure if they are enough for quantum.

1 answer

Last reply by: Professor Hovasapian
Tue Mar 24, 2015 5:02 PM

Post by shashikanth sothuku on March 24, 2015

hi prof,
21i+8i = 29i at 8:00

Complex Numbers

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Complex Numbers 0:11
    • Representing Complex Numbers in the 2-Dimmensional Plane
    • Addition of Complex Numbers
    • Subtraction of Complex Numbers
    • Multiplication of Complex Numbers
    • Division of Complex Numbers
    • r & θ
    • Euler's Formula
    • Polar Exponential Representation of the Complex Numbers
  • Example I 14:25
  • Example II 15:21
  • Example III 16:58
  • Example IV 18:35
  • Example V 20:40
  • Example VI 21:32
  • Example VII 25:22

Transcription: Complex Numbers

Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to take a little bit of a detour from the Quantum mechanics and we are going to talk about complex numbers.0004

Let us get started.0010

Complex numbers, we represent √-1 with an I, so I² = -1.0014

Complex numbers are represented this way.0026

We usually use the letter Z for a complex number and it is something like A + B I.0029

A is the real part of the complex number Z and B we call the imaginary part of the complex number Z.0038

Again, z is the standard variable just like X.0051

For real numbers, we use Z for complex numbers.0055

Geometrically, the complex numbers are represented in a 2 dimensional plane.0057

Geometrically, we represent complex numbers in 2 dimensional plane called the complex plane, with something like this.0066

This is the real axis, the real number line that you are used to.0101

And here we just call it the imaginary axis.0104

Again, this is just a pictorial way of actually representing a complex number.0107

If we have some complex number like this, let us say we mark a couple of points here 1,0111

this is -1 and on the imaginary axes this becomes +I and this becomes –I.0117

If I have some complex number like that, this is going to be the B this part and in this part right here that is going to be the A.0122

That is it, nice and simple.0142

Let us talk about some operations with complex numbers.0144

Our operations, you can add, subtract, multiply, and divide complex numbers the way you do any others.0147

Let us talk about addition first.0155

I think it is best represented with an example here.0160

Addition, if I have 2 + 3 I + 7 – 4 I, what you are going to do is you are going to add0163

the real part with the real part and you are going to add the imaginary part with the imaginary part.0175

It ends up being 2 + 7 = 9, 3 -4 = -1 I.0181

There you go, 9 – I, nice and simple.0190

Let us talk about subtraction, it is the same exact thing.0198

You are just going to be changing some signs.0202

Subtraction, let us do the same thing.0205

Let us do 2 + 3 I this time we will subtract 7 -4 I.0207

Once again, you are just going to put 2 -7 = -5 and you have 3 I—4 I = +7 I.0213

Nothing very strange about it.0226

Multiplication is also very straightforward.0228

For multiplication you are just going to treat it like you would any other binomial.0232

We have 2 + 3 I and we have the 7 -4 I so we are going to do this with this, the inside the outside, that same thing.0245

Or if you want this with this, this with this, and this with this.0256

How you do it actually does not matter.0259

2 × 7 we have 14 and we have to the inside we have + 21 I, we have -8 I, and here we have this is -3 × 4 = 12.0262

I × I = I² and I² = -1.0280

This becomes 14, 21 I -8 I = 13 I.0284

I hope you are checking my arithmetic, I'm notorious for being bad at arithmetic.0291

And this becomes + 12, this becomes 26 + 13 I.0296

Just multiply it out.0304

Before I do division, I want to introduce the notion of complex conjugate.0308

If A + B I is a complex number, then its complex conjugate which is denoted as Z*.0315

Z* = A – BI.0340

Basically, all you are doing is switching the + to A - .0344

Geometrically, it look something like this.0348

If this is the complex number, its conjugate is just a reflection along the x axis.0351

A + B I, A – B I, that is all it is.0359

Let us go ahead and do a division here.0365

Now, we have 2 + 3 I and we would divide that by the 7 -4 I.0370

We are going to multiply the top and bottom by the complex conjugate which is multiplying by 1.0378

We are going to end up multiplying the top and bottom.0385

7 -4 I the conjugate is 7 + 4 I / 7 + 4 I.0388

You have seen all this before but it is always nice to do a little bit of a review0394

because complex numbers are going to be very complex numbers and complex functions.0398

Very important in Quantum Mechanics.0401

We multiply the top we end up with 14, this is going to be + 21 I, this is going to be +8 I, and this is going to be + 12 I² ÷ 7 × 7 = 49.0406

This is going to be 4 × 4 this is going to be -16 I² and this is going to equal.0425

This is going to equal 14 + 39 I, this is +12 I², I² is -1 so it becomes -12.0433

Again, mostly this is just an arithmetic issue.0450

Keeping all the positives and negatives in order.0452

This is going to be I² is -1 so this is going to be 49 + 16 and we are going to get the 14 -12 =2 + 39 I ÷ 65.0456

Given a complex numbers in the form A + Bi this is how you handle the basic operations, very straightforward.0476

Let us go back to our geometrical representation.0486

Since, we have a geometrical representation in a 2 dimensional plane, this is our origin, this is our complex number.0490

We have this and we have this, this distance is A, this distance is B, this is the imaginary axis, this is the real axis.0500

Because of a point in the plane, we can also represent complex numbers in the polar form.0510

If this is an angle θ and we call this distance R, this R we represent as we call it just the modulus of the complex number.0517

Basically, it is just the length of the vector from 0 to the complex number.0534

I do not have to refer to it as a vector, it is just the length from the origin to that particular point.0540

It is exactly what you think, it is the Pythagorean Theorem.0548

It is A² + B² all over the radical.0550

R = we represent it that way.0562

Again, it is called the modulus of the complex number.0565

It is equal to A² + B² under the radical.0569

This Z when we squared, it is also equal to Z × Z conjugate.0577

Let me go back a little bit.0592

I have the Z, the R here that is equal to the A² + B².0596

Let us talk about θ, it is just the inverse tan of D/ A.0600

If you have R and θ, you can find A and B.0609

If you have A and B, you can find R and θ.0613

Let us go ahead and there is this relationship here which exists.0620

It is equal to Z × Z conjugate.0628

Again if you take the Z A + BI multiplied by the conjugate which is A – BI.0633

If you do that multiplication you can end up with A² + B²,0639

When you get rid of it when you square both sides, you will end up with this relation.0646

Reasonably important relation that tends to come up a lot.0653

Let us talk about Euler’s formula.0660

Euler’s formula says the following, E ⁺I θ = cos θ + I sin θ.0662

This is a very important formula.0681

Given R and θ, we can express A + BI in the form R × E ⁺I θ.0684

Basically, I have this version A + BI or I have this version of the complex number.0715

The relationship is as follows so A = R × the cos θ and B = R sin θ.0720

You have the relationships now.0734

Going one way, if you are given R and θ you can find A, R and θ you can find B.0736

You can go from the polar version to the Cartesian version.0740

Or if you are given A and B, you can use A and B to find R, and you can use the inverse tan of B and A to find θ.0742

You can go both ways.0752

This and this, and this and this, allows you to go back and fourth.0754

Let us take a look at how this happens.0762

A + BI, if A = R cos θ + I × R sin θ which is B.0765

I will go ahead and factor out the R.0781

I get R × cos θ + I sin θ.0783

Cos θ + sin θ = E ⁺I θ.0794

I have got R × E ⁺I θ.0798

That shows that this and this are equivalent.0802

There are different ways of actually writing the complex number.0808

Most of the time, this is going to be the most convenient form but occasionally this will be the most convenient form.0811

It just depends on what you are doing.0816

This is the polar exponential representation of the complex number.0821

If Z = RE ⁺Iθ.0829

The conjugate equals RE ⁻I θ.0835

Remember, positive angle negative angle, we are just reflecting it along the x axis.0842

This is very important E ⁺I θ, E ⁻I θ.0852

Let us go ahead and do some examples.0859

I think that is probably the best thing to do at this point.0862

Find the real and imaginary parts of 3 – 2 I².0866

Let us just do the math.0871

3 -2I² is going to be 3 – 2I.0873

We are going to get 3 × 3 = 9, this is going to be – 6I.0879

This is going to be -6I, this is going to be +4I².0889

We are going to get 9 -12 I – 4, 9 - 4 = 5, 5 -12 I.0895

The real part of Z = 5, the imaginary part of Z = -12.0906

That is it, very simple.0918

Find the real and imaginary parts of E⁻³ + I × π/ 3.0922

Let us go ahead and separate this out.0934

We have E⁻³ × this is -3 +,0936

The exponents are added which means 2 things with the same base are multiplied.0941

E⁻³ × E ⁺Iπ/ 3.0946

E⁻³ E ⁺I θ =cos θ + I × the sin θ.0953

Θ in this case is π/ 3.0962

We get E⁻³, the cos π/ 3, π/ 3 is 60°.0968

The cos of 60° is going to be ½ + I × 3/ 2 so we get E⁻³/ 2 + I × E⁻³ × 3/ 2.0975

Therefore, this is the real part and this is the imaginary part.1001

This is the real and this is the imaginary part.1007

You are just taking what is given and you are re representing it.1010

Express I – rad 3 in the form R E ⁺I θ.1020

We know what this is, R = we said it is A² + B² of the radical and we said that θ = the inverse tan of B/ A.1026

R = 5² + - rad 3² all over the radical that is going to be equal to 25 + 3 under the radical.1047

We have a √28 and θ = the inverse tan of - rad 3/ 5.1066

When I do that, I end up with a radium measure of -0.333.1081

Z = √28 × e⁰.333 × I.1093

There you go, RI of θ.1103

This is very straightforward as long as you have the back and forth relationship between the two.1110

Express e ⁻π/ 4 + LN 3 in the form A + BI.1119

Let us see what we have got.1127

I’m going to represent this as E ⁺Iπ/ 4 × E ⁺LN 3 which is equal to E ⁺LN 3 × cos of - I/ 4 + I × sin of -π/ 4.1132

Because here θ is -π/ 4.1161

This is equal to E × LN 3, cos -A = cos A.1165

First quadrant, the cos is positive.1174

This stays cos, this becomes the positive cos π/ 4.1177

The sin –π/ 4 is –sin π/ 4.1182

This turns into –I × sin π/ 4.1185

We get E ⁺LN 3, the cos π/ 4 is 1/ rad 2 – 1 × rad 2.1192

We know have E ⁺LN 3/ rad 2 – E ⁺LN 3/ rad 2 × I.1210

There is your A, there is your B, nice and simple.1231

Prove E ⁺Iπ =-1.1243

I just thought I would throw this in there.1246

It is one of all things that is really cool because you got the most important numbers and mathematics.1246

You got 1, you got e, you got I and π.1252

There is this really beautiful relationship among them.1256

Again, E ⁺Iπ you just represent it in sin cos form.1258

E ⁺Iπ= cos π + I × sin π.1265

The cos π is -1 and the sin π is 0, so that goes to 0.1277

That is it -1 and -1.1284

I just want you to see it.1286

The Euler’s formula is the cos θ + I sin θ ⁺nth = cos N θ + I × sin N θ.1296

Basically, just put the n in here.1307

Use this to prove the following trigonometric identity.1309

Sin 2 θ=2 sinθ cosθ.1312

You remember this trig identity from high school and we are going to prove it using Euler’s formula.1315

Let us see what we can do.1324

In this particular case, here we are going to let N=2.1326

This is going to equal to, we get cos θ + I × sin θ.1339

The reason that I pick N = 2 is because we are looking for sin 2 θ.1349

Sin 2 θ, I just was this thing here.1354

Let me just go ahead and set it equal to 2.1357

Cos θ + I sin θ², we need to show that it is actually equals cos 2 θ + I × sin 2 θ.1360

Let us go ahead and actually multiply this to prove that this is equal.1379

Let us work with this side and multiply it out.1384

I get cos² θ + I sin θ cos θ is I sin θ cos θ + 2I sin θ cos θ.1386

I’m just multiplying this out and then it is going to be + I² sin² θ.1400

This is going to be cos² θ + 2I sin θ cos θ.1410

I² is -1 so this is going to be -1 so this is going to be –sin² θ.1417

Let me put the real parts together.1421

I have cos² θ – sin² θ, that is that one.1424

I have +2i × sin θ cos θ.1430

I’m going to actually equate the real part of this with the real part of this and the imaginary part of this with the imaginary part of this.1439

I end up getting two identities.1466

I end up with cos 2 θ = cos² θ – sin² θ.1469

I get sin 2 θ= 2 sin θ cos θ.1479

Again, what we did was we equated real and imaginary parts on both sides.1487

We already know that the Euler’s theorem is roughly using this to prove the identity so there is no question mark as to whether they are equal.1503

And that is it, nice and straightforward.1514

Consider the following set of functions, the C sub N θ = 1/ 2π × e ⁺I n θ where N can be 0, + or -1, + or -2.1524

And θ is in the interval from 0 to 2π inclusive.1539

We want to show the following.1543

We have 4 things to show.1545

We want to show that the integral from 0 to 2π of this function.1546

The Z = 0 for all values of N that are not equal to 0.1552

The rad with 2π when N does equal 0.1558

We want to show that the integral of the Z conjugate × Z.1564

The integral of that equals 0, when N and M do not equal each other.1570

In other words, like 5 and 3 and.1575

The integral from 0 to 2π of that same integrand = 1 when N and M do equal each other.1578

Let us see what we can do.1586

Let us go ahead and start with this one fist.1588

Let me go ahead and do this in blue actually.1590

This is number 1, this is number 2, this is number 3, and this is number 4.1595

Let us do number 1.1603

For n not equal to 0, here is what we have.1606

The integral from 0 to π of Z sub N D θ, that is actually going to equal the integral from 0 to 2 π.1617

Here is our function, it is going to be, I’m going to pull the 1/ 2π out.1633

I’m just going to go ahead and write that as a constant.1641

It is going to be 1/ √2π × the integral from 0 to 2π of e ⁺I n θ D θ,1644

That is equal to 1/ 2π under the radical × the integral from 0 to 2π of cos θ + I sin θ.1661

I’m sorry this is N θ.1673

Just represent this using Euler’s formula, cos n θ + I sin N θ.1678

This is of the integral operator, the integration is actually linear so I can separate these out.1688

It was actually going to equal 1/ √2π × the integral from 0 to 2π of cos n θ D θ + 1/ the integral from 0 to 2π.1694

I’m going to put this I, I’m going to pull I out also.1718

Sin N θ D θ and this is equal to, this is 0 to 2π of the cos N θ.1723

This is 0 and this is 0.1733

It ends up equaling 0 which is what we said we have to prove.1737

That takes care of the first integral.1741

The second integral for N = 0.1745

Let us see, for N =0, Z sub N θ =1/ 2π is going to be e⁰ which is going to equal 1/ 2π rad.1750

We are going to have the integral from 0 to 2π of 1/ √2 π D θ.1779

That is going to equal 2π/ √2π.1793

This is just 2π/ 2π ^½ = 2π¹/2 which is √2π.1799

That takes care of N=0.1816

Let us see, for N not = M.1821

We said that the complex conjugate is just 1/ √2Π × E ⁻I N Θ, with the negative sign there.1832

The integral from 0 to 2π of the Z* × Z D θ is going to equal 1/ 2π now the radical sign goes away because it is going to be 1/ √2π × 1/ √2 π.1855

I’m multiplying them.1873

The complex conjugate of this function, this is the complex conjugate.1875

We will just go ahead and do this as is.1883

We have E ⁺I N Θ × E ⁺I M Θ D Θ.1886

That is going to equal 1/ 2 Π OF E ⁺I × Θ × M – N D Θ.1906

M-m is just some integer.1929

Because m is an integer and n is an integer.1940

What you end up getting is 1/ 2 π × the integral from 0 to 2 π of E ⁺I let us just call it P θ.1943

This is just the same as the one that we did before.1959

This ends up being 1/ 2 π × the integral from 0, this is going to be cos P θ + I sin P θ Dθ.1964

That is just going to equal 0.1986

That takes care of that one.1988

For our final one, we want to show that 1n = m.1992

Therefore, we are going to get Z* × Z = E ⁻I N Θ × E ⁺I N Θ = E ⁺I Θ N-N = E⁰.2003

We are going to have this integral from 0 to 2 π Z* × Z D θ.2032

When n = m, it is nothing more than 1/ 2 π × the integral from 0 to 2 π D θ,2041

That just equals 2 π/ 2 π which equals 1.2053

Thank you so much for joining us here at www.educator.com.2062

We will see you next time, bye.2064