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Consider P(z) = 2' + 2z-3 - i and, using the argument theorem and Rouche's theorem calculate the number of its roots in each of the following domains:(a) ...

Question

Consider P(z) = 2' + 2z-3 - i and, using the argument theorem and Rouche's theorem calculate the number of its roots in each of the following domains:(a) {:lz -1/ < 1;(b) (zIz - I/> 1lzl < 2} ,(c) {z/zl > 2/ .

Consider P(z) = 2' + 2z-3 - i and, using the argument theorem and Rouche's theorem calculate the number of its roots in each of the following domains: (a) {:lz -1/ < 1; (b) (zIz - I/> 1lzl < 2} , (c) {z/zl > 2/ .



Answers

The equation $\left(1-p^{2}\right) x^{2}+2 p x-1=0$ has both its roots in the interval $(0,2)$, if (a) $\mathrm{p}<-1$ (b) $-\frac{1}{2}<\mathrm{p}<1$ (c) $p>\frac{3}{2}$ (d) $\mathrm{p}>3$

You have our number 67 in which it is given that there is a function a practical to P X squared plus Q X square plus are okay? PX esquire Plus. Qx esquire affects equal to B X squared. Bless qx esquire Plus. Our what we say are expressed as our express as is satis satisfying the mean value there. Um and 0 to 1. Then we have to find the value of C in the interval 0-1. So I mean value terms to get the value of see we have according to Maine volunteer um after sequel to Yes for a B f B minus F A but B minus A. So this is the mean value to them Here. A zero B is one. And so let us first differentiate this. It will be to be X plus two Q x plus. Are we have to find it had X equal to see And now FB that F 1 -50 By 1 0. Right? Oh, let us right by plugging X equal to see Too busy plus two QC Plus are equal to F one value of function at one will be people ask you plus are processed T plus Q plus are pluses and minuses. Everything will be easy to accept us divide by 1 0. So no need to write anything as an as well get canceled out. Okay to here do be see plus two Q. C. And R. And r will also get canceled out. Mhm. So we'll be getting to P C plus two Q c minus P. My school equal to zero. So from these two legacy ps common to c minus one plus. From these two if they take us common It will be two C -1 equal to zero. Which means to see minus when into people's Q equal to zero. Either they should be equal to zero or two. C minus one should be equal to zero, so C equal to one by two, so sequel to one by two should be the correct answer, and option number eight is the correct option. Thank you.

To solve the problem. The following is given a phoenix equal to be x squared plus cubic square. Please. Are the knicks plus s on on the closing double zero comma one. Now we have to find the value of the sea in the opening double zero comma one such that the functional faces satisfies the magnitude. That is the value of F dash C will be equally into FNB -5. A. Over B -AB is a human to one and easy here in 20. That is if dash C will be equal in two Fn one minus infant zero. What? One minus zero. Now let us first evaluate the value of a flash eggs to find what the value of a dash C. Now the F dash X is equal into twice of PX place twice of Q X. Blessed art. The same applies that the value of a taxi will be twice of bill C plus twice of qc plus are subsiding the value of taxi in equation one will obtain the following. That is twice a pc bliss price of Q. C plus R is equal to F 11 -30. That is myself Pc place twice of Qc place are is 80 million two P plus Q. Yes our this is minus is so this implies that twice a pc plus twice of Qc is equal into, he plays Q. The same place that she is willing to people skew over twice of people skew that is the value of C Is equal into one x 2. What we can see that the correct option is option E. one way to is the value of C. Which belongs to the opening double zero comma one.

We are trying to see if where we're always stare um applies, We do need to show that first of all that it does apply since this is a polynomial function. The function itself is continuous on the interval from 0 to 2. And since it's a polynomial function, it is also differentiable everywhere. And specifically it is differential on the interval Open in the role from 0 to 2. 3rd condition is that the function at the end points Must be the same function is zero would be 0 0 -2. So negative too. And the function at two, if we fail to in forex would be four -4 -2, which is also negative too. By roistering the derivatives should be zero. Somewhere in the interval Are derivative is two x minus two. Now where is that equal to 0? Just a little two step equation and the two And divide by the two. So it's true when X is one. Mhm

Thank. This problem is asking us to discuss what happens if we take the two routes from this equation and multiply those two routes together specifically wants us to prove that if we take our first and our second route and multiply them together, we will always get see over a well. First off, we would need to define what are how we represent the roots of this equation. Recognize that a X squared plus BX plus C is a quadratic equation, And the easiest way that I know of a lease to represent roots of a quadratic equation is using the quadratic formula right. The quadratic formula tells us that if we have a quadratic equation in the form of a X squared plus BX plus C, then X is going to be equal to negative B plus or minus the square root of B squared, minus four a c all over to a So that is our two routes right there, right? One of them is the plus version. One of them is the minus version. So for argument's sake, let's say that we say Route one is the plus version, so Route one is negative B plus B squared minus for a C all over two. A. And let's say then that route to is the negative version. So it's negative. B minus B squared, minus four a. C all over two. A. If we think of it this way, then what they're really asking us to do is to take these two expressions and multiply them together. Are one Times are to root root one times route to. And if we multiply these together, hopefully we will get see over a. If we did it correctly, right? That's what they're looking for us to get. So if we take negative B plus B squared minus four a C all over two a times negative B minus B squared, minus four a c all over two A. Well, that would mean that we're multiplying fractions, right? If we're multiplying fractions, the way we're supposed to do that is you take the two numerator and multiply them together and you take the two denominators and you multiply them together. Our denominators are pretty straightforward to eight times to a well, that's just for a squared. All right. Two times two is 48 times a is a squared for our numerator. Recognize that these are quantities, right? We have a negative B and A B squared minus for a C in our numerator. So this is going toe have to be a foiling problem. All right, that's what's happening here. Meaning we wanna look at this as we're taking negative B so specifically, the highlighted negative B. And we want to distribute that to both of the terms in our numerator. Negative B times negative B would be a positive B squared negative B times negative B squared, minus four a. C under the square root negative times a negative is a positive, and since B is outside the radical and B squared minus four, A. C is inside the radical. We can't really combine them. We just have be square root of B squared minus four a. C, which is a really terrible looking term. But it's not a problem because now we need to take the B squared the square root of B squared minus four a c, the one that's highlighted in blue, and we need to distribute that as well. And if you take the square root of B squared minus four a. C. times a negative. B you get negative B square root of B squared minus for a C, which means these two middle terms air going to cancel out. All right. Finally, we take the square root of B squared minus four a C times the negative square root of B squared, minus four a. C. A positive times. A negative is a negative. And then, if you're taking a square root of B squared minus four a C times a square root of B squared minus four A. C, you are effectively taking the square root of B squared minus four a. C squared. You're taking it times itself. So then, looking at this are two middle terms here are going to cancel, so that's all taken care of. And then a square root being squared also cancels, meaning we are left with B squared minus B squared, minus for a C, the quantity right because the square in the square root cancel each other out. And then finally, that is all over for a squared. Still, we want to distribute that negative so weakened. Drop the parentheses so a negative B squared would be, or a negative times B squared is negative B squared and then a negative times a negative for a C is a positive for a C. We'll certainly be squared minus B squared. Cancels leaving us with four A c over four. A squared before is would cancel. You have a single A and eight the first in your numerator and to a c A squared in the denominator, Which means simplifying that would get rid of this a on top and leave you with only a single A in the denominator giving you see over a as a final answer. So that would be your proof that this works pretty long. The hardest part is definitely this foiling that you have to do in the numerator. All of this stuff right here is probably the toughest to get from here to this step. But if you can get past that, I think the rest of it kind of solves itself down in the end. And this is your answer to see over a


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