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JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Easiest way to proceed is the square both sides and see what you get from there

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1so ... waiting for you to take a step or engage in the conversation

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0couldnt 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

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

Yahoo!
 2 years ago
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

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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Squaring 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?

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Or is z here a complex number?

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1If z is complex, we're going to have use a different approach.

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1right. So in that case, what have you tried?

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Write 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.

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

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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, now see if you can reduce this somehow

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360344301541:dw

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, 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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360344512672:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360344710864:dw damn this is getting complicated.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360344933595:dw

sirm3d
 2 years ago
Best 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

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

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.3oops, 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)
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