If #|Z - 4/Z| = 2#, then the maximum value of #|Z|# is equal to?

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Answer 1 · 2018-02-06T14:28:24

#max absZ = sqrt(2 (3 + sqrt[5])) = 1+sqrt5#

Assuming that #Z in CC# we have

#Z = rho e^(i phi)# and then knowing that

#absZ = sqrt(Z bar Z)# follows

#(rho e^(i phi)-4/rho e^(-i phi))(rho e^(-i phi)-4/rho e^(i phi)) = 4# or

#rho^2-8cos(2phi)+16/rho^2 = 4# or

#rho^4-2(2+4cos(2phi))rho^2 +16=0#

Now solving for #rho^2#

#rho^2 = 2 (1 + 2 Cos(2 phi) pm sqrt[4 Cos(2 phi) + 2 Cos(4 phi)-1])#

for #phi = 0# we have

#rho_(max)^2 = 2 (1 + 2 + sqrt[4 + 2 -1]) = 2(3+sqrt(5))#

then

#max absZ = sqrt(2 (3 + sqrt[5])) = 1+sqrt5#