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Ican be shoan thatUse thbs Iacl (and I04 operatons) = Ind *, and "z Ihal satisty Ihe equation'Shoon AnsIro51n6CnereinsuiknClneclicck Nisatr...

Question

Ican be shoan thatUse thbs Iacl (and I04 operatons) = Ind *, and "z Ihal satisty Ihe equation'Shoon AnsIro51n6CnereinsuiknClneclicck Nisatr

Ican be shoan that Use thbs Iacl (and I04 operatons) = Ind *, and "z Ihal satisty Ihe equation 'Sho on Ans Iro51n6 Cnereinsuikn Clne clicck Nisatr



Answers

In Exercises 55–58, you may use a graphing calculator to solve the problem.
$(\mathbf{A}) z+\overline{z} \quad(\mathbf{B}) z \overline{z}(\mathbf{C})(z+\overline{z})^{2} \quad(\mathbf{D})(z \overline{z})^{2}(\mathbf{E}) z^{2}$

For this problem were given to complex number Z one and Z two. And a is asking us to multiply Z one and Z two and leave it in the uh in the polar form. So we're gonna go ahead and multiply Z one is 7 82, Degree I. Times Z two is to E. To the 31 degree I. So we know that when we're multiplying, okay the numbers in front of you can go ahead multiply them together seven times two is 14. And for the exponents we know that when we multiply multiply for example, X squared times X. To the cube. We know that when we multiply we always add the exponents. So this is the same thing as X. To the fifth power. So you would do the same thing here, you will add the exponents here. So you will figure out how 80 to 82 I plus 31. Eye is going to be how many eyes. So 82 plus 31 that's going to give you, so that's going to eat to the 113. 1313 degree I. And that's your answer. Okay, we just combine those two things together and therefore part B is asking us to divide them so we can divide, so again Z one is seven E 82 to 82 degree I. Over the two is 2 E. To the 31 Degrees I. So when you divide again, you can divide seven over to 7/2. I'm just gonna leave as a fraction 7/2. And then when you're dividing with exponents, let's say X. The sixth over X to the fourth, we know that we should subtract those two exponents, so six minus four, that's gonna be X squared. So you will do the same thing here, take your exponents 80 to 82 degree I. And now you're going to subtract 31 degree high because you're dividing and 80 to subtract 31, that's going to give you 51 Degrees i. And that's the answer for part B.

In this problem, we want to show that X plus y plus Z is the local linear approximation to the function X times y Times Z plus two at the specified point and also that two x minus y minus C plus two is a local linear approximation. Four x all over Why plus Z again at the same point. So to do so, we'll use the fact that the local linear approximation of a function of three variables at a point P is equal to F evaluated at P plus the partial derivative of F with respect to X evaluated at p times a quantity X minus The X coordinate of P plus the partial derivative of F with respect toe wine at p times y minus the y coordinate of p finally plus the partial derivative of F with respect to Z evaluated at p times a quantity Z minus the z coordinate of the point. So starting with part A, we have the function x times y times Z plus two. So let's start by taking the partial derivative of F with respect to X, remember that we treat X as the only variable and when we do that. We get that this is equal to Y Times Z and then the partial derivative of F with respect to why that's equal to X. Times E in the partial derivative of F with respect to Z is equal to x times. Why and since we have the 0.1 comma, one comma one. All of these partial derivatives will be equal at that point because it's just one times one for each case. So we get that the local linear approximation to our function is f evaluated at our point, which would be one times one times one plus two. So that's three plus the partial derivative of F with respect to X at P, which is one times of quantity X minus one plus one times a quantity. Why minus one plus one times the quantity Z minus one. We get that that is equal to X plus why plus C and then for part B will continue in the same way. So let's find the partial derivative of F with respect two X, And for this function we have four x all over y plus Z. So the partial derivative of F with respect to x iss four over y plus e and this time, the partial derivatives derivatives at the point p that's given they won't all be the same. So let's go ahead and evaluate those, uh, each one at a time, and we get when we plug in our point. We get that this is for over two, which is just to the partial derivative of F with respect toe y that is equal to minus four x times a quantity. Why plus E to the minus two. And we want to evaluate this at our point, and we get that that's minus four all over two squared, which is minus one. And then we want to take the partial derivative of F with respect to Z, and we get that That's minus for X times, the quantity y plus C to the minus two. And then we want to evaluate this at our point. And of course, we get minus one. And so now we just want to plug these values into our local linear approximation. But first, our function for ex all over white plus e evaluated at the point is for over two, which is just to plus two times the quantity X minus one minus the quantity. Why minus one minus the quantity Z minus one, which is equal to two plus two acts minus two minus Why plus one minus e plus one. And then from here, these twos will cancel. And we get that. This is two acts minus why minus c plus two, which is thesafeside thing that we have above, and that completes the problem.

For this problem we are given two complex numbers which is which are in the polar form were asked to use those two complex numbers to multiply them out Z one time C two and divide them Z one over Z two. So and we want to leave our answer in the poll form at the end as well so we can go ahead and bring them down and multiply them. So we know Z1 is 3.05 E. to the 1.76 i. Times Z two is 11.94 e. to the 2.59 i. So the numbers in front of my we can just multiply them together so you can use your calculator uh 3.05 times 11.94. That's going to give you 36.417. And because the exponents have the same base which is E we know that when we're multiplying which she add the exponents so we can add what's 1.76 I plus 2.59 irs. And that's going to give you four point 35 I. And that's the answer for part A. In the polar form and part B we're going to do the same thing. So we know what Z one equals two which was 3.5 E. To the 1.76 I Oversee to we know it was 11 .94 e. to the 2.59 i. So again when we're dividing we can just use our calculator to divide what's 3.05 divided by 11.94. Okay, so you can live as a decimal, but I'm going to leave leave as a fraction. And that's the same thing as 305 over Uh 1194. And then because the exponents have the same base which is E so when we're dividing, we subtract the exponents, so you're going to do 1.76, subtract 2.59, which is gonna be negative 0.83 then just bring the eye down. So that's the answer for part B.

Hell in this problem we have given Z even is equal to 3.5 in two, he told the bubble 1.76 iota on Zito is equals toe 11.94 in two a to the power 2.59 iota Onda were asked to finds even into Zito. Ondas, even a Ponzi to In the Polar Form, was even into Zito. We will multiplies even and Zito. We get 3.5 in two e to the power 1.76 iota and to 11.94 into E to the power 2.59 i o Town. Now we will are the Tita because they both have iota. So we get 3.5 in 2. 11.94 is equal to 3.5 in 2. 11.94 is 36.417 into he to the power 1.76 plus 2.59 iota is 4.35 i o town now for the even Upon Zito, we will divides even an Zito. We get 3.5 in two e to the bubble 1.76 iota upon 11.94 into he to the power 2.59 iota. Now whether subjected Tita now whether subjected Tita because they both have iota. So we get 3.5 upon 11.94 is 0.255 into here to the power of 1.76 iota minus 2.59 iota is minus 0.8 iota. Hence vegan Z one into Zito is equals toe 36.417 in two e to the power 4.35 into one iota on Z even upon Zito is equals toe zero going toe 55 and two e to the power minus 0.8 iota.


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