Question

If Amp $\left(\frac{z+2}{z-4i}\right)=\frac{\pi }{2}$ then the locus of Z = x+iy is

• A. $x^2+y^2+3x-2=0,=0$
• B. x = 0 such that y > 2/3
• C. $x^2+y^2-2x-4y=0suchthaty<0x^2+y^2>1,$
• D. $x^2+y^2+2x-4y=0suchthat2x-y+4>0$

Answer 1

Answer: (D)

$x^2+y^2+2x-4y=0suchthat2x-y+4>0$

$\begin{array}{c}wehave,\\ Amp\left(\frac{z+2}{z-4i}\right)=\frac{\pi }{2}andlocusofz=x+iy\\ Now,\\ complexno.isz=\left(\frac{z+2}{z-4i}\right)=pureimagenaryno.\\ then,\\ z=x+iy\\ \Rightarrow \frac{(x+2)+iy}{x+(y-4)i}\times \frac{x-(y-4)i}{x-(y-4)i}\\ \Rightarrow Z_1=\frac{x(x+2)+y(y-4)+i\left[yx-(x+2)(y-4)\right]}{x^2+{(y-4)}^2}\\ Rl\left(Z_1\right)=0\\ x(x+2)+y(y-4)=0\\ \therefore x^2+y^2+2x-4y=0,\\ \Rightarrow yx-(x+2)(y-4)>0\\ \Rightarrow yx-xy-2y+4x+8>0\\ \therefore 2x-y+4>0\\ So,bothconditionaresatisfiedandthecorrectoptionisD.\end{array}$