Need help in sketching complex inequations (1 Viewer)

[Vipul_K]

 

[Run hard@thehsc]

@Vipul_K Just plug in x = iy and then solve for o. In part (a) you solve for the imaginary part to be zero and then sketch that graph and in part (b) you solve for the real part to be zero and then sketch that graph. Make sure to exclude any points by looking at the domain of the graph... (Correct me if I am wrong ppl)....

 

[Vipul_K]

@Vipul_K Just plug in x = iy and then solve for o. In part (a) you solve for the imaginary part to be zero and then sketch that graph and in part (b) you solve for the real part to be zero and then sketch that graph. Make sure to exclude any points by looking at the domain of the graph... (Correct me if I am wrong ppl)....

YES thank you! I did exactly that earlier but I kept messing up! but now I have gotten it! thanks!

 

[icycledough]

So as mentioned above, I think the only restriction would be z = i (so 1 unit above the origin on the y-axis), due to the denominator being zero. So you should just be able to put an open circle there, and go about it how you normally would.

 

[CM_Tutor]

If is real, then the locus of has the ratio of the moduli as a real constant, giving a circle where the line joining the two fixed points (here and ) is divided internally and its projection divided externally in the ratio of the constant to 1. The centre of the circle lies on this line and one of the two points is inside the circle and the other is outside.