Arg(2z + z bar)
= arg(2x+2yi+x-yi)
= arg(3x+yi)
= tan-1(y/3x)
=
/4 [Given]
therefore, y/3x = 1
or, y = 3x
Now,
= mod(x + yi + 3 + 4i) + mod(x + yi + 2 + i)
= mod[(x+3) + (y+4)i] + mod[(x+2) + (y+1)i]
= mod[(x+3) + (3x+4)i] + mod[(x+2) + (3x+1)i]
=
(x
2 + 6x + 9 + 9x
2 + 24x + 16) +
(x
2 + 4x + 4 + 9x
2 + 6x + 1)
=
(10x
2 + 30x + 25) +
(10x
2 + 10x + 5)
= 5 [
(2x
2 + 6x + 5) +
(2x
2 + 2x + 1) ]
Now we have to find the minimum value of the variables under the root...
Inorder to find that, we convert them to perfect square.
The minimum value of perfect square is 0.
=
5 [
{ (
2x + 3/
2)
2 + 1/2 } +
{ (
2x + 1/
2 )
2 + 1/2 } ]
=
(5/2) [
{ (2x+3)
2 + 1} +
{ (2x+1)
2 + 1} ]
Here the minimum value will not be 0 for both of them at any one particular valueof 'x'.
The minimum value of the two squares will exist at -1.
Thus, the minimum value will be:
= 2
5
Moderation Team
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