Question

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

Answer 1

\[\left| \frac{Z^2+9}{Z} \right|=6\]
top part always positive
\[Z^2+9=6\left| Z \right|\]
\[Z^2-6\left| Z \right|+9=(Z-3)(Z-3)\]

Answer 2

\[(Z-3)^2=0\]
Z=3 since Z has no other valueZ=3

Answer 3

Lol....i tried this way.......the answer shuld be 3 + sqrt (18)

Answer 4

since\[\left| Z \right|=\sqrt{Z^2}=a\]
\[Z^2+6\sqrt{Z^2}+9=0\]
\[a-6a^2+9=0\]
\[6a^2-a-9=0\]

Answer 5

@satellite73

Answer 6

i made a wrong substitution\[a^2-6a+9=0\]
\[a=3\]
\[Z=\pm3\]

Answer 7

using calculus method we derive\[Z^2+9=6\left| Z \right|\]
\[2Z=6\frac{Z}{|Z|}\]
\[|Z|=3\]

Answer 8

Isn't answer 27 ?

Answer 9

the answer shuld be 3 + sqrt (18)

Answer 10

ohh I mean sqrt27
hmm, still now getting.

Answer 11

not*

Answer 12

ok...any how Can u show Ur Steps ?

Answer 13

I got messed up. Sorry
Gimme some time.

Answer 14

@mukushla

Answer 15

Let z=x+iy
| (x+iy)^2 + 9| = 6sqty(x^2 + y^2)

|(x^2 -y^2 + 9) + i(2xy)| = 6sqrt(x^2 + y^2)

=>(x^2 -y^2 + 9)^2 + (2xy)^2 = 36(x^2 + y^2)

We need to maximize (x^2 + y^2) with the given info that
(x^2 -y^2 + 9)^2 + (2xy)^2 = 36(x^2 + y^2)

Yes the answer is 3 + sqrt18