# If z=x+i*y prove that |x|+|y|=<(square root 2)*|z|

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### 3 Answers

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|]**