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\[Z+ \frac{ 9 }{ Z } = 6\] ,Then find the Greatest value of Z ?
 one year ago
 one year ago
\[Z+ \frac{ 9 }{ Z } = 6\] ,Then find the Greatest value of Z ?
 one year ago
 one year ago

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JamesJBest ResponseYou've already chosen the best response.1
Easiest way to proceed is the square both sides and see what you get from there
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
so ... waiting for you to take a step or engage in the conversation
 one year ago

funinaboxBest ResponseYou've already chosen the best response.0
couldnt you also just say Z + 9/Z = 6, and ignore the absolute value, because you are looking for the largest value anyways then solve for Z as: Z^2 + 9 = 6z Z^2  6x + 9 = 0 and go from there
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
that gives you some of the solutions. You also need to consider the case z + 9/z = 6
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@JamesJ that would give answer as Z = 3.... But the answer shuld be Z = 3 + sqrt18
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Note: Z is a Complex Number
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Squaring both sides \[ \left z + 9/z \right^2 = 6^2 \] i.e., \[ (z + 9/z)^2 = 36 \] as x^2 = x^2 for all real numbers x. Hence \[ z^2 + 18 + 81/z^2 = 36 \] and therefore \[ z^4  18z^2 + 81 = 0 \] Now you have a quadratic equation in z^2 which you can solve... \[ (z^2  9)^2 = 0 \] \[ z^2 = 9 \] \[ z = 3 \] yes. The other answer you just wrote down doesn't satisfy the original equation, does it?
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Or is z here a complex number?
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
If z is complex, we're going to have use a different approach.
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
right. So in that case, what have you tried?
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Write z = x +iy as usual. Now notice the original equation is equivalent to z^2 + 9 = 6z Now write out those two expressions in terms of x and y and see what you get. It will be easier to square them.
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
i.e., write out \[ z^2 + 9^2 = 36 z^2 \] for example, \[ z^2 = x^2 + y^2 \] You take the next step ....
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
(x^2 y^2 + 9)^2 + (2xy)^2 = 36(x^2 + y^2)
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Yes, now see if you can reduce this somehow
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1360344301541:dw
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
Yes, it's not clear to me that we get much from the polar approach. Where do you go from there? Have you studied Lagrange multipliers btw? This problem looks a lot like a Lagrangian problem. You have a constraint in x and y given by this equation, and you want to maximize f(x,y) = x^2 + y^2
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Nope..i havent Studied that
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1360344512672:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1360344710864:dw damn this is getting complicated.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1360344933595:dw
 one year ago

sirm3dBest ResponseYou've already chosen the best response.3
\[6=\leftZ+\frac{9}{Z}\right>\leftZ\left\frac{9}{Z}\right\right\]put \[z=re^{i\theta}\\z=r,\quad \left\frac{9}{z}\right=\frac{1}{r}\] now solve the inequality in r
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
since we have isolated the value r as function of theta we might maximize it using simple derivatives. but this is not really suggested.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.3
oops, that's 9/r r9/r<6 r9/r < 6 r^2  9 < 6r r^2  6r < 9 r^2  6r + 9 < 18 (r3)^2 < 18 (r3) < sqrt(18) r< 3 + sqrt(18)
 one year ago
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