locus question (1 Viewer)

[Accuracy]

find locus of z if (z-i) / (z+ 1) is purely real

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{Let}~z=x@plus;iy\\ \frac{z-i}{z@plus;1}\\ =\frac{x@plus;i(y-1)}{(x@plus;1)@plus;iy}\\ =\frac{[\left x@plus;i(y-1)\right ][\left(x@plus;1)-iy ]\right}{[\left(x@plus;1)@plus;iy]\right [\left(x@plus;1)-iy ]\right}\\ ~\\ \textrm{If a complex number is purely real its imaginary part is 0 i.e.}\\ -xy@plus;(y-1)(x@plus;1)=0\\ -xy@plus;xy@plus;y-x-1=0\\ y=x@plus;1\\ ~\\ \therefore \textrm{The locus of z is a straight line with gradient 1 with x-intercept (-1,0) and y-intercept (0,1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{Let}~z=x+iy\\ \frac{z-i}{z+1}\\ =\frac{x+i(y-1)}{(x+1)+iy}\\ =\frac{[\left x+i(y-1)\right ][\left(x+1)-iy ]\right}{[\left(x+1)+iy]\right [\left(x+1)-iy ]\right}\\ ~\\ \textrm{If a complex number is purely real its imaginary part is 0 i.e.}\\ -xy+(y-1)(x+1)=0\\ -xy+xy+y-x-1=0\\ y=x+1\\ ~\\ \therefore \textrm{The locus of z is a straight line with gradient 1 with x-intercept (-1,0) and y-intercept (0,1)}" title="\\ \textrm{Let}~z=x+iy\\ \frac{z-i}{z+1}\\ =\frac{x+i(y-1)}{(x+1)+iy}\\ =\frac{[\left x+i(y-1)\right ][\left(x+1)-iy ]\right}{[\left(x+1)+iy]\right [\left(x+1)-iy ]\right}\\ ~\\ \textrm{If a complex number is purely real its imaginary part is 0 i.e.}\\ -xy+(y-1)(x+1)=0\\ -xy+xy+y-x-1=0\\ y=x+1\\ ~\\ \therefore \textrm{The locus of z is a straight line with gradient 1 with x-intercept (-1,0) and y-intercept (0,1)}" /></a>