In Exercises 77–80, convert to polar form and then perform the in...

Question

In Exercises 77–80, convert to polar form and then perform the indicated operations. Express answers in polar and rectangular form.
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(1 + i√3)(1 − i) / 2√3 − 2i

Accepted Answer

Welcome back. I am so glad you're here. We're asked to express the given number in polar form, simplify the given expression and write your answer in polar and rectangular form. Our given complex number is in the numerator, we have the quantity of the square it of three plus three, I multiplied by the quantity of four minus four I. And that is all divided by the quantity of five minus five I square at three. So we have a lot of steps here. The first thing we want to do is write this number in polar form. So we recall from previous lessons that polar form is given with R multiplied by the quantity of the cosine of theta plus I sine theta and R is our modulus. And theta is our argument. We find R because that is equal to the square ret of A squared plus B squared. And we find theta because the tangent of theta equals B divided by A. And how this links back is that we know when we have a complex number that is given to us as A plus B I. So here we have three complex numbers with two different operations happening in the numerator. The operation between two complex numbers is multiplication. And then between the numerator and the denominator, we have the operation of division. So these three complex numbers first have to be turned into polar form, all three of them. And then we'll do, we'll simplify and we'll do our operations. Then we stated in both polar and rectangular form. So let's start off by taking each of these three complex numbers in the numerator, we're going to start with the square it of three plus three I. And here we can identify our A and our B. So A is the first number not attached to an I A. Here is the square of three B is the one being multiplied by the I. Here B is a positive three. Now we can start filling that in into sulfur R and for theta. So R equals the square of, instead of A, we have the square of three that needs to be squared plus instead of B, we have three that needs to be squared. The squared of three squared is three and three squared is nine. So in the radical, we have three plus not 12, 3 plus nine, which is 12. So we have the square of 12 which simplifies 22 square at three. So our R for this first complex number is two square at three. Now let's figure out what theta is. So we know that the tangent of theta equals B, we said B is three divided by A is the square of three. All right, we don't want to have that radical in the denominator. So we'll simplify this by multiplying both the numerator and the denominator by the square of three. That's going to be three square at three divided by three, which simplifies to simply the square of three. So we know that the tangent of theta equals the square it of three. Where is this true? Well, this is true in quadrant one. And we're going to stay in quadrant one because both our B and our A are positive. We know that B or excuse me, let's start with A, we know that A is the horizontal component and we know that B is the vertical component. So when they're both positive, we're in the first quadrant. So when is the tangent of theta equal to the square of three in quadrant one? That is going to be when theta equals 60 degrees. So we'll stay in degrees for our polar form. So writing inside of the parentheses, we had two square three multiplied by the quantity of. Now in the parentheses, we have the cosine of not theta but 60 degrees plus I sign of not theta but degrees. OK. We got that one written in polar form. We have to do the same thing for the other two complex numbers. So I'll go a little bit more quickly here. So we have four minus four, I, for our second complex number, we see that our A here is a positive four and B is a negative four. And now we can plug those in to sol for R and theta. So R equals the square at of A is four squared plus B is negative four squared, four squared is 16 plus negative four squared is also 16, 16 plus 16 is 32 and 32. That or excuse me, the square rut of 32 is going to simplify to be four square root two. So our R here is four square foot two. Now let's find theta. We know that the tangent of theta equals B which is negative four divided by A which is positive four, negative four divided by four simplifies to negative one. So we have the tangent of theta equals negative one. Where is this true? Well, tangent is negative in quadrant two and quadrant four. But as we said before, the horizontal component, our A here is positive and our vertical component B is negative, that's gonna put us down in the fourth quadrant. And so if we were thinking of a reference angle in quadrant one, saying that the tangent of theta equals positive one in quadrant one, our data equals pi divided by four. And using that as our reference angle in quadrant four, we would have theta equal to seven pi divided by four. But since we're in degrees instead of radiance seven pi divided by four is 315 degrees. So we have four square root two multiplied by the quantity of the cosine of 315 degrees plus I sine of 315 degrees. There is the second one. Now the third one in the denominator, we have five minus five I square at three. I'm going to write this as minus five square at three. And then the I just noting that the I is being multiplied by negative five square at three, but it's not inside the radical with the three. So our A here is a positive five and our B here is negative five square at three. We see we're going to be with the positive horizontal component and the negative vertical component again. So again, in quadrant four, let's figure out R R equals the square root of a square. That's five squared plus B squared. B is negative five square at three squared. So that's still in the radical. We have 25 plus multiplied by three, which is 75 and 25 plus 75 is 100. The square root of 100 is a positive 10. So our R here is positive 10. Now to figure out the, we know, yeah, let's put some lines in here so that this doesn't all go together. There we go. OK. The tangent of theta equals B divided by A B is negative five square. At three divided by A is positive five and the fives cancel out. So this just simplifies to a negative square at three. And so very similar to the first one where we had the tangent of theta equals positive square at three. But now we know we're down in the fourth quadrant as we mentioned before. And so we know that in the first quadrant, this was 60 degrees. If we go down to the fourth quadrant, we can take 360 minus 60 degrees. That's going to give us a positive degrees. So we have 10 multiplied by the quantity of the cosine of degrees plus I sine of degrees. OK. So now let's rewrite this in polar form. So in the numerator, we've got two square at three multiplied by the quantity of cosine 60 degrees plus I sine 60 degrees. And that whole thing is being multiplied by four square with two multiplied by the quantity of cosine 315 degrees plus I sine of 315 degrees. And that whole numerator is being divided by 10, multiplied by the quantity of the cosine of 300 degrees plus I sine 300 degrees. OK. So we reroute that from, from its complex number into polar form. Now, our steps are to do the multiplication and the division to simplify this have our answer in polar form and rectangular form smell. Let's move to a different part of the screen where we'll have more room to work. So in order to multiply complex numbers in polar form, what we're going to do is multiply the module. So in the numerator, the module are two square at three and four square foot two. So we'll start with that two square at three, multiplied by four square root two is eight square at six. And now we're going to add the arguments, the arguments here are 60 degrees plus 315 degrees. And that's going to give us 375 degrees. So we have eight square foot six multiplied by the quantity of the cosine of 375 degrees plus I sine degrees. All still divided by 10 multiplied by the quantity of cosine 300 degrees plus I sine 300 degrees. OK. Now, to divide the second operation here, to divide complex numbers in polar form, we are going to divide the module. The module here are eight square foot six and 10. And then we are going to subtract the arguments 375 degrees and 300 degrees. So eight square at six divided by 10 simplifies two, four square at six, divided by five. And that's all multiplied by the quantity of degrees minus 300 degrees is 75 degrees. So cosine degrees plus I sign 75 degrees. And so that is our simplified answer in polar form. Now we need to get this into rectangular form. Now let me, let me highlight that one as one of our answers and highlight this one as our answer for it in polar form before we simplify it. OK. So we rewrote it in polar form. Now, we have simplified it in polar form now to get it to rectangular form. So what we wanna do is we want to take four square at six divided by five and we want to multiply it by the exact values for the cosine of 75 degrees and the sign of degrees. And for the cosine of 75 degrees, you can put that into your calculator, that one's not on the unit circle and you'll get the square it of six minus the square of two all divided by four and just make sure you're in degrees. And then we can bring down plus I and then for the sign of 75 degrees, we'll get the square out of six plus the square out of two all divided by four. And now we just need to distribute this four square root six divided by five to the square at six minus square at two, all divided by four and the square at six plus square at two all divided by four. And when we do that, we get six minus two square at three, all divided by five plus. And I'm going to put the number in front of the I, we'll have six plus two square at three all divided by five, not multiplied by the I. So that one is in standard form, there is our simplified answer in rectangular form. All right. Well done. We'll catch you on the next one.

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In Exercises 77–80, convert to polar form and then perform the indicated operations. Express answers in polar and rectangular form. _ _ (1 + i√3)(1 − i) / 2√3 − 2i