4) Put z = lzlcis  ;

so LHS = lcis -1l = 2*l sin /2 l * l i*cis /2 l..........{also, l i*cis /2 l = 1 } ;

            = 2*lsin /2l ;

            <= l  l = l arg(z) l.............ans.

 

3) given :  x+i(y+2)+*[(x-1)²+y²] = 0........

also,given tht  is a real no...

so, y = -2 ; x+*[(x-1)²+y²] = 0 ;

i.e; x²(²-1) - 2x² + 5 = 0 ;

x is real...so, discriminant >=0.........

⁴-5² + 5 >= 0 ...u can find range of  frm this.......ans.