WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.
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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.
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Let us get started.
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Complex numbers, we represent √-1 with an I, so I² = -1.
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Complex numbers are represented this way.
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We usually use the letter Z for a complex number and it is something like A + B I.
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A is the real part of the complex number Z and B we call the imaginary part of the complex number Z.
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Again, z is the standard variable just like X.
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For real numbers, we use Z for complex numbers.
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Geometrically, the complex numbers are represented in a 2 dimensional plane.
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Geometrically, we represent complex numbers in 2 dimensional plane called the complex plane, with something like this.
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This is the real axis, the real number line that you are used to.
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And here we just call it the imaginary axis.
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Again, this is just a pictorial way of actually representing a complex number.
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If we have some complex number like this, let us say we mark a couple of points here 1,
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this is -1 and on the imaginary axes this becomes +I and this becomes –I.
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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.
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That is it, nice and simple.
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Let us talk about some operations with complex numbers.
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Our operations, you can add, subtract, multiply, and divide complex numbers the way you do any others.
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Let us talk about addition first.
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I think it is best represented with an example here.
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Addition, if I have 2 + 3 I + 7 – 4 I, what you are going to do is you are going to add
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the real part with the real part and you are going to add the imaginary part with the imaginary part.
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It ends up being 2 + 7 = 9, 3 -4 = -1 I.
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There you go, 9 – I, nice and simple.
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Let us talk about subtraction, it is the same exact thing.
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You are just going to be changing some signs.
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Subtraction, let us do the same thing.
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Let us do 2 + 3 I this time we will subtract 7 -4 I.
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Once again, you are just going to put 2 -7 = -5 and you have 3 I—4 I = +7 I.
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Nothing very strange about it.
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Multiplication is also very straightforward.
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For multiplication you are just going to treat it like you would any other binomial.
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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.
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Or if you want this with this, this with this, and this with this.
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How you do it actually does not matter.
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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.
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I × I = I² and I² = -1.
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This becomes 14, 21 I -8 I = 13 I.
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I hope you are checking my arithmetic, I'm notorious for being bad at arithmetic.
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And this becomes + 12, this becomes 26 + 13 I.
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Just multiply it out.
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Before I do division, I want to introduce the notion of complex conjugate.
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If A + B I is a complex number, then its complex conjugate which is denoted as Z*.
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Z* = A – BI.
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Basically, all you are doing is switching the + to A - .
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Geometrically, it look something like this.
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If this is the complex number, its conjugate is just a reflection along the x axis.
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A + B I, A – B I, that is all it is.
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Let us go ahead and do a division here.
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Now, we have 2 + 3 I and we would divide that by the 7 -4 I.
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We are going to multiply the top and bottom by the complex conjugate which is multiplying by 1.
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We are going to end up multiplying the top and bottom.
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7 -4 I the conjugate is 7 + 4 I / 7 + 4 I.
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You have seen all this before but it is always nice to do a little bit of a review
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because complex numbers are going to be very complex numbers and complex functions.
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Very important in Quantum Mechanics.
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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.
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This is going to be 4 × 4 this is going to be -16 I² and this is going to equal.
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This is going to equal 14 + 39 I, this is +12 I², I² is -1 so it becomes -12.
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Again, mostly this is just an arithmetic issue.
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Keeping all the positives and negatives in order.
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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.
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Given a complex numbers in the form A + Bi this is how you handle the basic operations, very straightforward.
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Let us go back to our geometrical representation.
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Since, we have a geometrical representation in a 2 dimensional plane, this is our origin, this is our complex number.
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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.
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Because of a point in the plane, we can also represent complex numbers in the polar form.
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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.
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Basically, it is just the length of the vector from 0 to the complex number.
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I do not have to refer to it as a vector, it is just the length from the origin to that particular point.
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It is exactly what you think, it is the Pythagorean Theorem.
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It is A² + B² all over the radical.
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R = we represent it that way.
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Again, it is called the modulus of the complex number.
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It is equal to A² + B² under the radical.
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This Z when we squared, it is also equal to Z × Z conjugate.
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Let me go back a little bit.
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I have the Z, the R here that is equal to the A² + B².
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Let us talk about θ, it is just the inverse tan of D/ A.
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If you have R and θ, you can find A and B.
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If you have A and B, you can find R and θ.
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Let us go ahead and there is this relationship here which exists.
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It is equal to Z × Z conjugate.
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Again if you take the Z A + BI multiplied by the conjugate which is A – BI.
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If you do that multiplication you can end up with A² + B²,
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When you get rid of it when you square both sides, you will end up with this relation.
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Reasonably important relation that tends to come up a lot.
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Let us talk about Euler’s formula.
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Euler’s formula says the following, E ⁺I θ = cos θ + I sin θ.
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This is a very important formula.
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Given R and θ, we can express A + BI in the form R × E ⁺I θ.
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Basically, I have this version A + BI or I have this version of the complex number.
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The relationship is as follows so A = R × the cos θ and B = R sin θ.
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You have the relationships now.
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Going one way, if you are given R and θ you can find A, R and θ you can find B.
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You can go from the polar version to the Cartesian version.
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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.
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You can go both ways.
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This and this, and this and this, allows you to go back and fourth.
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Let us take a look at how this happens.
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A + BI, if A = R cos θ + I × R sin θ which is B.
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I will go ahead and factor out the R.
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I get R × cos θ + I sin θ.
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Cos θ + sin θ = E ⁺I θ.
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I have got R × E ⁺I θ.
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That shows that this and this are equivalent.
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There are different ways of actually writing the complex number.
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Most of the time, this is going to be the most convenient form but occasionally this will be the most convenient form.
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It just depends on what you are doing.
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This is the polar exponential representation of the complex number.
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If Z = RE ⁺Iθ.
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The conjugate equals RE ⁻I θ.
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Remember, positive angle negative angle, we are just reflecting it along the x axis.
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This is very important E ⁺I θ, E ⁻I θ.
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Let us go ahead and do some examples.
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I think that is probably the best thing to do at this point.
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Find the real and imaginary parts of 3 – 2 I².
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Let us just do the math.
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3 -2I² is going to be 3 – 2I.
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We are going to get 3 × 3 = 9, this is going to be – 6I.
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This is going to be -6I, this is going to be +4I².
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We are going to get 9 -12 I – 4, 9 - 4 = 5, 5 -12 I.
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The real part of Z = 5, the imaginary part of Z = -12.
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That is it, very simple.
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Find the real and imaginary parts of E⁻³ + I × π/ 3.
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Let us go ahead and separate this out.
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We have E⁻³ × this is -3 +,
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The exponents are added which means 2 things with the same base are multiplied.
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E⁻³ × E ⁺Iπ/ 3.
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E⁻³ E ⁺I θ =cos θ + I × the sin θ.
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Θ in this case is π/ 3.
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We get E⁻³, the cos π/ 3, π/ 3 is 60°.
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The cos of 60° is going to be ½ + I × 3/ 2 so we get E⁻³/ 2 + I × E⁻³ × 3/ 2.
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Therefore, this is the real part and this is the imaginary part.
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This is the real and this is the imaginary part.
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You are just taking what is given and you are re representing it.
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Express I – rad 3 in the form R E ⁺I θ.
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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.
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R = 5² + - rad 3² all over the radical that is going to be equal to 25 + 3 under the radical.
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We have a √28 and θ = the inverse tan of - rad 3/ 5.
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When I do that, I end up with a radium measure of -0.333.
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Z = √28 × e⁻⁰.333 × I.
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There you go, RI of θ.
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This is very straightforward as long as you have the back and forth relationship between the two.
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Express e ^-π/ 4 + LN 3 in the form A + BI.
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Let us see what we have got.
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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.
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Because here θ is -π/ 4.
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This is equal to E × LN 3, cos -A = cos A.
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First quadrant, the cos is positive.
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This stays cos, this becomes the positive cos π/ 4.
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The sin –π/ 4 is –sin π/ 4.
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This turns into –I × sin π/ 4.
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We get E ⁺LN 3, the cos π/ 4 is 1/ rad 2 – 1 × rad 2.
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We know have E ⁺LN 3/ rad 2 – E ⁺LN 3/ rad 2 × I.
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There is your A, there is your B, nice and simple.
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Prove E ⁺Iπ =-1.
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I just thought I would throw this in there.
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It is one of all things that is really cool because you got the most important numbers and mathematics.
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You got 1, you got e, you got I and π.
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There is this really beautiful relationship among them.
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Again, E ⁺Iπ you just represent it in sin cos form.
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E ⁺Iπ= cos π + I × sin π.
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The cos π is -1 and the sin π is 0, so that goes to 0.
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That is it -1 and -1.
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I just want you to see it.
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The Euler’s formula is the cos θ + I sin θ ⁺nth = cos N θ + I × sin N θ.
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Basically, just put the n in here.
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Use this to prove the following trigonometric identity.
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Sin 2 θ=2 sinθ cosθ.
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You remember this trig identity from high school and we are going to prove it using Euler’s formula.
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Let us see what we can do.
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In this particular case, here we are going to let N=2.
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This is going to equal to, we get cos θ + I × sin θ.
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The reason that I pick N = 2 is because we are looking for sin 2 θ.
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Sin 2 θ, I just was this thing here.
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Let me just go ahead and set it equal to 2.
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Cos θ + I sin θ², we need to show that it is actually equals cos 2 θ + I × sin 2 θ.
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Let us go ahead and actually multiply this to prove that this is equal.
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Let us work with this side and multiply it out.
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I get cos² θ + I sin θ cos θ is I sin θ cos θ + 2I sin θ cos θ.
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I’m just multiplying this out and then it is going to be + I² sin² θ.
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This is going to be cos² θ + 2I sin θ cos θ.
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I² is -1 so this is going to be -1 so this is going to be –sin² θ.
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Let me put the real parts together.
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I have cos² θ – sin² θ, that is that one.
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I have +2i × sin θ cos θ.
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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.
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I end up getting two identities.
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I end up with cos 2 θ = cos² θ – sin² θ.
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I get sin 2 θ= 2 sin θ cos θ.
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Again, what we did was we equated real and imaginary parts on both sides.
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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.
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And that is it, nice and straightforward.
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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.
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And θ is in the interval from 0 to 2π inclusive.
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We want to show the following.
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We have 4 things to show.
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We want to show that the integral from 0 to 2π of this function.
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The Z = 0 for all values of N that are not equal to 0.
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The rad with 2π when N does equal 0.
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We want to show that the integral of the Z conjugate × Z.
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The integral of that equals 0, when N and M do not equal each other.
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In other words, like 5 and 3 and.
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The integral from 0 to 2π of that same integrand = 1 when N and M do equal each other.
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Let us see what we can do.
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Let us go ahead and start with this one fist.
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Let me go ahead and do this in blue actually.
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This is number 1, this is number 2, this is number 3, and this is number 4.
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Let us do number 1.
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For n not equal to 0, here is what we have.
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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.
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Here is our function, it is going to be, I’m going to pull the 1/ 2π out.
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I’m just going to go ahead and write that as a constant.
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It is going to be 1/ √2π × the integral from 0 to 2π of e ⁺I n θ D θ,
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That is equal to 1/ 2π under the radical × the integral from 0 to 2π of cos θ + I sin θ.
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I’m sorry this is N θ.
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Just represent this using Euler’s formula, cos n θ + I sin N θ.
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This is of the integral operator, the integration is actually linear so I can separate these out.
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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.
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I’m going to put this I, I’m going to pull I out also.
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Sin N θ D θ and this is equal to, this is 0 to 2π of the cos N θ.
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This is 0 and this is 0.
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It ends up equaling 0 which is what we said we have to prove.
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That takes care of the first integral.
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The second integral for N = 0.
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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.
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We are going to have the integral from 0 to 2π of 1/ √2 π D θ.
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That is going to equal 2π/ √2π.
00:29:59.600 --> 00:30:15.800
This is just 2π/ 2π ^½ = 2π¹/2 which is √2π.
00:30:15.800 --> 00:30:21.500
That takes care of N=0.
00:30:21.500 --> 00:30:32.500
Let us see, for N not = M.
00:30:32.500 --> 00:30:55.100
We said that the complex conjugate is just 1/ √2Π × E ⁻I N Θ, with the negative sign there.
00:30:55.100 --> 00:31:13.300
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.
00:31:13.300 --> 00:31:15.100
I’m multiplying them.
00:31:15.100 --> 00:31:23.300
The complex conjugate of this function, this is the complex conjugate.
00:31:23.300 --> 00:31:26.500
We will just go ahead and do this as is.
00:31:26.500 --> 00:31:46.500
We have E ⁺I N Θ × E ⁺I M Θ D Θ.
00:31:46.500 --> 00:32:09.100
That is going to equal 1/ 2 Π OF E ⁺I × Θ × M – N D Θ.
00:32:09.100 --> 00:32:20.600
M-m is just some integer.
00:32:20.600 --> 00:32:23.500
Because m is an integer and n is an integer.
00:32:23.500 --> 00:32:38.700
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.
00:32:38.700 --> 00:32:43.700
This is just the same as the one that we did before.
00:32:43.700 --> 00:33:05.800
This ends up being 1/ 2 π × the integral from 0, this is going to be cos P θ + I sin P θ Dθ.
00:33:05.800 --> 00:33:08.400
That is just going to equal 0.
00:33:08.400 --> 00:33:11.800
That takes care of that one.
00:33:11.800 --> 00:33:22.700
For our final one, we want to show that 1n = m.
00:33:22.700 --> 00:33:51.900
Therefore, we are going to get Z* × Z = E ⁻I N Θ × E ⁺I N Θ = E ⁺I Θ N-N = E⁰.
00:33:51.900 --> 00:34:01.500
We are going to have this integral from 0 to 2 π Z* × Z D θ.
00:34:01.500 --> 00:34:13.000
When n = m, it is nothing more than 1/ 2 π × the integral from 0 to 2 π D θ,
00:34:13.000 --> 00:34:22.100
That just equals 2 π/ 2 π which equals 1.
00:34:22.100 --> 00:34:23.800
Thank you so much for joining us here at www.educator.com.
00:34:23.800 --> 00:34:25.000
We will see you next time, bye.