On line and so on Given that Z2 is a complex number, | 2i-z2 | = 1, why does it mean that the trajectory of Z2 is a circle with radius 1 and center (0,2)?

On line and so on Given that Z2 is a complex number, | 2i-z2 | = 1, why does it mean that the trajectory of Z2 is a circle with radius 1 and center (0,2)?

z=x+iy
|2i-z2|=│-x-i(y-2)│=√[x²﹢(y﹣2)²]＝1
That is X & # 178; + (Y-2) &# 178; = 1
The trajectory of ∧ Z2 is a circle with radius 1 and center (0,2)

Complex problems in Senior Two
Let a belong to R, z = x + Yi, x, y belong to R, known Z & # 178; - A & # 178; / Z & # 178; + A & # 178; is a pure imaginary number, the conditions that x, y should meet (kneel down to a mathematical master, explain the steps in detail, urgent!

z=x+yiz^2=x^2-y^2+2xyiz^2-a^2=(x^2-y^2-a^2)+2xyiz^2+a^2=(x^2-y^2+a^2)+2xyi (z^2+a^2)(z^2-a^2)=(x^2-y^2+a^2)^2+4x^2y^2 (z^2-a^2)^2=(x^2-y^2-a^2)^2-4x^2y^2+4xy(x^2-y^2-a^2)i(x^2-y^2-a^2)^2-4x^2y^2=0x^2-...

Will the equation with imaginary coefficients have real roots?
If not
Example: if the equation x square + (K + 2I) x + 2 + ki = 0 of X has a real root, then the value of the real number k is_____ ?
Root sign 2
Substituting the value of K into the coefficient is still an imaginary number. Why is there a real root?
To find the explanation Tat - (* ＾ - ゜) V

x²+(k+2i)x+2+ki=0
(x²+kx+2)+(2x+k)i=0
x²+kx+2=0
2x+k=0
have to
x=±√2
k=±2√2