In Exercises 53–64, use DeMoivre's Theorem to find the indicat...

Question

In Exercises 53–64, use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. (√2 − i)⁴

Accepted Answer

Welcome back. I am so glad you're here. We are asked to find the quantity of square at three minus I square at three race to the fifth power using Demo's theorem. Express the answer in rectangular form. Our answer choices are answer choice. A negative 36 square at three plus 36 I square at three. Answer choice B negative 36 plus 36 I answer choice C negative nine plus 36 I square at two and answer choice D 18 minus nine I square rot three. All right. So the first thing we want to do in order to utilize Demos theorem is to rewrite this part that's inside of the parentheses. Kind of ignoring that it's raised to the fifth power for a moment. In polar form, we recall from previous lessons. The polar form is when we have R multiplied by the quantity of the cosine of theta plus I sine theta. And we recall from previous lessons that R equals the square rut of A squared plus B squared. And to find theta, we know that the tangent of theta equals B divided by A. And so where are A and B coming from, well, A and B, when we've got this in rectangular form are going to be, it's going to be A plus B I. So A is the number that's not attached to the I. Here A equals the square of three and B is going to be the number that is attached to the I. So here B equals the negative square of three. Let's find R first, we know R equals the square of A squared plus B squared A is the squared of three. And we'll square that plus B is negative square at three and we'll square that. Well, the square root of three squared is just three and the negative square root of three squared is also three. So we're taking the square rot of three plus three and three plus 36. So this is the square root of six for our R. Now let's work on theta. And we said that the tangent of theta equals B divided by A. So we said B is negative square at three and A is the square of three while taking the negative square out of three divided by the square of three equals a negative one. So we're saying that the tangent of theta equals negative one, but which quadrant are we in? Well, we know that A is the horizontal component. So A here is positive, that means we're going to be the, to the right of the Y axis and B is the vertical component here B is negative. So that means we're going to be below the X axis and that means we're in the fourth quadrant. So thinking for a moment for our reference angle, if we were talking about the tangent of beta being equal to a positive one and quadrant one, that would be for the angle pi divided by four. And so in quadrant four, the corresponding angle there is going to be seven pi divided by four. All right. So filling that in, we've got the square out of six multiplied by the quantity of the cosine of seven pi divided by four plus I multiplied by the sign of seven pi divided by four. All right, we've got that in polar form. Now, going back to Demo's theorem, we know that Z raised to the end power equals R raised to the end power multiplied by the quantity of the cosine of N theta plus I sine N theta. So what's our N? Our N is what our quantity is being raised to. It's raised to the fifth power. So N equals five. Now, we can fill that in. So we've got Z raised to the fifth power equals RR is the square of six. And we're raising that to the fifth power that's multiplied by the quantity of the cosine of N. Here is five, multiplied by theta is seven pi divided by four plus. And then I need a little bit more room with this, I multiplied by the sign of an RN is five multiplied by beta, which is seven pi divided by four. All right. So now let's simplify that. So we've got the squared of six raised to the fifth power. Well, that's going to be six multiplied by itself five times inside of the radical. And for two of those, or for four of them, they'll get to come out as two sixes. So this will be 36 square at six, multiplied by the quantity of the cosine of five multiplied by seven pi is 35 pi divided by four plus I sine again 35 pi divided by four. Now we can simplify 35 pi divided by four to get that within from 0 to 2 pi we'll find a co terminal angle. So subtracting our two pi and when we are in with a denominator of four, that's going to be subtracting eight pi divided by four. So if we have 35 pie, that's not a 35. If we have 35 pie minus eight pie, that brings us down to 27 pie, we subtract another eight pie that gets us to pie. Subtract another eight pie that gets us to 11 pie. Subtract another eight pie and that will bring us to the pie. So the co terminal angle that we can figure out for cosine is three pi divided by four. So taking 36 square re six multiplied by the quantity of the cosine of three pi divided by four plus I sine of three pi divided by four. And now we can recall from previous lessons, the cosine of three pi divided by four and the sign of three pi divided by four. So three pi divided by four, if we draw a quick coordinate plain horizontal X axis, a vertical y axis three pi divided by four is in quadrant two. And we know that from previous lessons, the exact value for the cosine of three pi divided by four is going to be square at two divided by two. And when we're in quadrant two cosine is negative. So we have a negative square root two divided by two plus I multiplied by now for the exact value for the sign of three pi divided by four. Again, the sign of three pi divided by four is also square root two, divided by two. Sine is positive and quadrant two. So that's gonna be a positive square at two divided by two, all still being multiplied by 36 square at six. The next step is to distribute this 36 square at six to the parentheses. And then we'll have our answer in rectangular form. So taking 36 square at six, multiplied by negative square at two divided by two, that is going to simplify to be negative 36 square at three. And then we'll have 36 square at six multiplied by a positive square at two, divided by two. And that one's going to be so we bring down plus I and then also 36 square it three. Now the only difference between the way we've written it and the answer choice is they put the 36 before the I, so that it is closer to standard form, they keep the square of three after it. So that we know that the I is not inside of the radical. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

In Exercises 53–64, use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form. (√2 − i)⁴

Key Concepts

DeMoivre's Theorem

Conversion Between Rectangular and Polar Forms

Rectangular Form of Complex Numbers

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