IIT JEE , AIEEE Forum - Mathematics , Algebra

rtiit

For every real number c>=0, find all the complex numbers z which satisfy the equation

2(mod z)-4cz+1+ic=0

rtiit

ne1 plz

Greatdreams

z = x + iy

mod z =

( x² + y²)

so 2

( x² + y²) - 4 c (x + iy) + 1+ ic = 0

equating real n imaginary parts,

- 4cy + c = 0

so y = 1/4

now put it in the eqn

you get 2

(x² + 1/16 ) - 4cx + 1 = 0

now you have 4 ( x² + 1/16) - 8cx + 3/4 = 0
solve for x ,

x = 8c

(64 c² - 12(4c² - 1) / 8 ( 4 c² - 1)

c is given

0

put the values of y and you get the complex number