|z-i|/|z+1|=√2. Magnitude of complex number=√(real2+imaginary2). It's assumed that the vertical bars denote magnitude, not the absolute value.
Let z=rcosθ+irsinθ≡a+ib (a=rcosθ, b=rsinθ), then:
|z-i|=|rcosθ+i(rsinθ-1)|=√(r2cos2θ+r2sin2θ-2rsinθ+1)=√(r2-2rsinθ+1);
|z+1|=|rcosθ+1+irsinθ|=√(r2cos2θ+2rcosθ+1+r2sin2θ)=√(r2+2rcosθ+1).
|z-i|/|z+1|=√((r2-2rsinθ+1)/(r2+2rcosθ+1))=√2, therefore:
r2-2rsinθ+1=2(r2+2rcosθ+1)=2r2+4rcosθ+2,
r2+2r(2cosθ+sinθ)+1=0,
There are two variables but only one equation so there are many solutions.
When θ=0, this becomes r2+4r+1=0,
r2+4r=-1,
r2+4r+4=3,
(r+2)2=3,
z=r=-2±√3 (real but irrational and negative) appears to be a solution.
CHECK SOLUTION:
z=√3-2:
|z-i|=|√3-2-i|=√(8-4√3)=2√(2-√3); |z+1|=|√3-1|=√3-1.
2√(2-√3)/(√3-1)=(√3+1)√(2-√3).
[(√3+1)√(2-√3)]2=2(2+√3)(2-√3)=2(4-3)=2. Therefore |z-i|/|z+1|=√2.
z=-√3-2:
|z-i|=|-√3-2-i|=√(8+4√3)=2√(2+√3); |z+1|=|-√3-1|=-√3-1.
2√(2+√3)/(-√3-1)=-2√(2+√3)/(√3+1)=-(√3-1)√(2+√3).
[-(√3-1)√(2+√3)]2=2(2-√3)(2+√3)=2(4-3)=2. Therefore |z-i|/|z+1|=√2.
Both solutions check out.
But these are only two out of many solutions.
When 2cosθ+sinθ=0, that is, when 2+tanθ=0, tanθ=-2, sinθ=-⅖√5, cosθ=⅕√5; or sinθ=⅖√5, cosθ=-⅕√5. In each case r2=-1, implying r is imaginary; but r has to be real. So the equation r2+2r(2cosθ+sinθ)+1=0 puts constraints on r, as shown below.
r2+2r(2cosθ+sinθ)=-1,
r2+2r(2cosθ+sinθ)+(2cosθ+sinθ)2=(2cosθ+sinθ)2-1,
(r+2cosθ+sinθ)2=(2cosθ+sinθ)2-1≥0,
4cos2θ+4sinθcosθ+sin2θ≥1,
3cos2θ+cos2θ+4sinθcosθ+sin2θ≥1,
3cos2θ+4sinθcosθ≥0,
(3cosθ+4sinθ)cosθ≥0. When θ=(2n+1)π/2, cosθ=0, but tanθ→∞.
When tanθ=-¾, 3cosθ+4sinθ=0, so nπ+tan-1(-¾)≤θ≤nπ+½π, where n is an integer.
Between these limits for θ, r is positive and z=rcosθ+isinθ, which is always complex except when θ=πn.
Therefore we can generate many values for z which satisfy the given equation. It can be shown that r cannot exceed √5+2 or be less than -√5-2, and there is no value for r between 2-√5 and √5-2.