If at least one value of a complex no z=x+yi satisfy the condition mod(z+root2)=root(a^2-3a+2) and the inequality mod(z+iroot2)< a then (a)a<2 (b) a>2 (c) a=2 (d) none Share with your friends Share 0 Lovina Kansal answered this Dear student Correct question isIf atleast one value of a complex number z=x+iy satisfy the condition z+2=a2-3a+2 and the inequality z+i2<a2 then(a) a<2 (b) a>2 (c) a=2 (d) noneSolution:If z=x+iy is a complex number satisfying the given conditions,thena2-3a+2=z+2⇒a2-3a+2=z+i2+2-i2⇒a2-3a+2≤z+i2+21-i ∵z1+z2≤z1+z2⇒a2-3a+2<a2+2⇒-3a<0⇒a>0Since z+2=a2-3a+2 represents a circle with centre at A-2,0and radius a2-3a+2 and z+2i<a2 represents the interior of the circlewith centre at B0,2 and radius a.Therefore, there will be a complex number satisfying the given condition and the given inequality if the distance AB is less than the sum or difference of the radii of the two circles i.e. if -2-02+0+22<a2-3a+2±a⇒2±a<a2-3a+2⇒4+a2±4a<a2-3a+2⇒-a<-2 or 7a<-2⇒a>2 or a<-72But a>0, ∴a>2 Regards 1 View Full Answer
