Question

if z= x+iy where x,y are real prove that |x|+|y|=<√2|z|​

Answer 1

Step-by-step explanation: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 -...