Answer to Question #288816 in Complex Analysis for Mos

Question #288816

Geometric representation of S={Im(z-i)=|z+I|}


1
Expert's answer
2022-01-19T18:10:26-0500

1. Let z=x+iy,x,y∈Rz=x+iy, x, y \in \R

Then


Im(z−i)=Im(x+iy−i)=y−1Im(z-i)=Im(x+iy-i)=y-1

∣z+i∣=∣x+iy+i∣=x2+(y+1)2|z+i|=|x+iy+i|=\sqrt{x^2+(y+1)^2}

We have the equation

y−1=x2+(y+1)2y-1=\sqrt{x^2+(y+1)^2}

x2+(y+1)2≥0=>y≥1\sqrt{x^2+(y+1)^2}\geq0=>y\geq1

If y≥1,y\geq1, then

x2+(y+1)2≥y+1>y−1\sqrt{x^2+(y+1)^2}\geq y+1>y-1

Therefore, the equation


y−1=x2+(y+1)2y-1=\sqrt{x^2+(y+1)^2}

has no solution.


2. Let z=x+iy,x,y∈Rz=x+iy, x, y \in \R

Then


Im(z−i)=Im(x+iy−i)=y−1Im(z-i)=Im(x+iy-i)=y-1

∣z+1∣=∣x+1+iy∣=(x+1)2+y2|z+1|=|x+1+iy|=\sqrt{(x+1)^2+y^2}

We have the equation

y−1=(x+1)2+y2y-1=\sqrt{(x+1)^2+y^2}

(x+1)2+y2≥0=>y≥1\sqrt{(x+1)^2+y^2}\geq0=>y\geq1

If y≥1,y\geq1, then

(x+1)2+y2≥y>y−1\sqrt{(x+1)^2+y^2}\geq y>y-1

Therefore, the equation


y−1=(x+1)2+y2y-1=\sqrt{(x+1)^2+y^2}

has no solution.



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