For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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### 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
- Complex Numbers
- 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
- Example II
- Example III
- Example IV
- Example V
- Example VI
- Example VII

- 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

### Physical Chemistry Online Course

### 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 add*0163

*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 review*0394

*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

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