Animated Solution for Mathematics - Complex Numbers: If $z=x+iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$, then

When approaching complex number problems, visualize the Argand plane. We are dealing with two distinct geometric constraints that define the location of $z$.

Our first condition is $|z|-2=0$, which simplifies to $|z|=2$. In the complex plane, the modulus $|z|$ represents the distance of a point $z$ from the origin $(0,0)$.

This equation defines a circle centered at the origin with a radius of $2$. Substituting $z=x+iy$, we obtain the Cartesian form:

The second condition is $|z-i|-|z+5i|=0$. Rearranging this gives $|z-i|=|z-(-5i)|$.

This represents the locus of all points $z$ equidistant from the points $(0,1)$ and $(0,-5)$. Geometrically, this is the perpendicular bisector of the segment joining these two points.

To find the equation of this line, we substitute $z=x+iy$:

Squaring both sides yields:

The $x^2$ terms cancel out, leaving us with $(y-1{)}^2=(y+5{)}^2$. Expanding this results in:

Simplifying the equation, we find $-12y=24$, which gives the horizontal line:

We now have two constraints: the circle $x^2+y^2=4$ and the line $y=-2$. To find the intersection, we substitute $y=-2$ into the circle equation:

This leads to $x^2=0$, which implies $x=0$. The only point satisfying both conditions is $(0,-2)$, corresponding to the complex number $z=-2i$.

To conclude, we verify the result against the provided options. Substituting $x=0$ and $y=-2$ into the equation $x+2y+4=0$:

The point satisfies the equation perfectly. The geometric perspective confirms that the intersection of these two loci is the singular point $z=-2i$.