3 Answers | Add Yours
We have to prove that |x|+|y|=<sqrt 2|z|, if z=x+i*y.
Now assume that the given relation is true. So |x|+|y|=<sqrt 2|z|.
Taking the square of both the sides
=> (|x|+|y|)^2 =< [(sqrt 2)*|z|]^2
=> x^2 + y^2 + 2|x|*|y| =< 2*|z|^2
As |z| = sqrt (x^2 + y^2)
=> x^2 + y^2 + 2|x|*|y| =< 2* (x^2 + y^2)
=> x^2 + y^2 + 2|x|*|y| =< 2 x^2 + 2 y^2
=> x^2 + y^2 - 2|x|*|y| >= 0
=> ( |x| - |y| )^2 >= 0
Now this is always true as the square of a term is always greater than 0.
Therefore our assumption that |x|+|y|=<sqrt 2*|z| is validated.
z = x+iy.
Therefore |z| = |x+iy| = sqrt(x^2+y^2) , where x and y are real numbers.
Pur r = x^2+y^2 .
Then cost = x/r
Sint sint y/r.
Therefore z = r (cost+isint).
|z| = r. ,
|x| = r|cost| and |y| = r|sint|.
|x|+|y| = r |cost|+r|sint| = < r *Max (sint +cost).....(1)
We know sint +cost is maximumum when (sint+ccost) = 0 for which t . (sint+cost)" < 0.
(sint+cost)' = o gives: cost-sint = 0 when x = pi/4 or t = -5pi/4.
And (sint+cost)'' = -(sint +cost) < 0 only for pi/4.
Therefore Max(sint+cost) = 1/sqrt2+1/sqrt2 = 2/sqrt2 = 2^(1/2).
Substituting Max (sint+cost) = sqrt2 in (1) we get:
|x|+|y| < r* 2^(1/2) = (sqrt2)*|z|.
Therefore |x|+|y| < = (sqrt2)|z|.
We'll raise to square both sides:
(|x|+|y|)^2=<[(square root 2)*|z|]^2
(|x| + |y|)^2 = x^2 + y^2 + 2|x|*|y|
[(square root 2)*|z|]^2 = 2*|z|^2
But |z| = sqrt[Re(z)^2 + Im(z)^2], where Re(z) = x and Im(z) = y
|z| = sqrt(x^2 + y^2)
We'll raise to square:
|z|^2 = x^2 + y^2
x^2 + y^2 + 2|x|*|y| =< 2(x^2 + y^2)
We'll subtract x^2 + y^2 + 2|x|*|y| both sides and we'll use symmetric property:
2(x^2 + y^2) - (x^2 + y^2 + 2|x|*|y|) > = 0
We'll remove the brackets and we'll get:
2x^2 + 2y^2 - x^2 - y^2 - 2|x|*|y| >= 0
We'll combine like terms:
x^2 + y^2 - 2|x|*|y| >= 0
(|x| - |y|)^2 >= 0 true, so
|x|+|y|=<[(square root 2)*|z|]
We’ve answered 318,982 questions. We can answer yours, too.Ask a question