IIT JEE , AIEEE Forum - Mathematics , Algebra

sachin_gupta1991

Q. If modz<=1, mod w<=1,

show that mod(z-w)² <= (modz-modw)² + [arg (z) - arg(w)]²

feynmann

Let z1 = r1 exp( i @1 )

z2 = r2 exp ( i @ 2 )

then we have r1 < = 1 & r2 < = 1

Now mod ( z1 - z2 ) ^2 = r1^2 + r2^2 -2r1 r2 cos ( @1 - @ 2 )

= ( r1 - r2 )^2 + 2r1r2 { 1 - cos (@1 _ @2 ) }

= (r1 - r2 ) ^2 + 4r1 r2 sin^2 ( @1 - @2 ) /2

Now we have r1, r2<= 1

& for the principal value of arguments , sin^2x<= x ^2

combining these results we get

mod( z1 - z2 )^2 <= (mod z1 - mod z2 ) ^2 +( arg z1 - arg z2 ) ^2

(proved)