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 +...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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.**