Find the locus of points in the plane satisfying each of the given conditions:
(i) |z-5| = 6 (ii) |z-2i| >=1 (iii) Re(z+2) = -1
(iv) Re{i(conjugate of z)} =3 (v) |z+i| = |z-i|
Also Sketch its diagram.
Please tell me as soon as possible...?
Locus Problem in complex number....?
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Answers (1)
My english is very bad but I will tell you. 1)z=x+yi Ix+yi-5I=6, (x-5)^2+(y)^2=36=the general equation of a circle. So Circle's Center is (5,0), r=6. Draw it.
2)z=f+gi If+i(g-2)I>=1 f^2+(g-2)^2=1=the general equation of a circle. So Circle's Center is (0,2), r=1 Draw it but Scan the outside of the circle.
3)z=a+bi R(z)=a a+2=-1 a=-3 from -3 Draw a perpendicular to the real axis.
4) z=c+di conjugateofz=c-di i*conjugateofz=ic+d Re(ic+d)=3 d=3 from +3 Draw a perpendicular axis to the complex.
5) z=m+ni Im+i(n+1)I=Im+i(n-1)I, m^2+(n+1)^2=m^2+(n-1)^2, m^2+n^2+2*n+1=m^2+n^2-2*n+1, 4n=0, n=0 means the real axis. You don't need to draw.