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I solved the problem correctly but my answer is rather long

|z-(-8i)|=4 (drawing)

z_0=a+bi, z_1=-a+bi

|a+bi+8i|=4

sqrt(a^2+(b+8)^2)=4 (radius)

a^2+b^2+16b+64=16 (eq 1)

And from 4-8-sqrt(48) triangle

sqrt(a^2+b^2)=sqrt(48)

a^2+b^2=48 (eq 2)

Place into eq 1

48+16b+64=16

16b=-96

b=-6

Placing into eq 2

a^2+36=48

a^2=12

a=+-sqrt(12)

z_0=sqrt(12)-6i

z_1=-sqrt(12)-6i

From the drawing, the smallest argument is at the third zone.

tan theta=-6/-sqrt(12)

=sqrt(3)

so theta=240 degrees.

There

**must**be a shorter solution, can you please help me?