Given that z is a complex number and the module of (z-2i) = 1, then the value range of the imaginary part of Z is

Given that z is a complex number and the module of (z-2i) = 1, then the value range of the imaginary part of Z is

Let z = a + bi, then (z-2i) = a + (b-2) I
From the fact that modules are equal to one, we know: A2 + (b-2) 2 = 1
In the coordinate system, the above formula can be regarded as x2 + (b-2) 2 = 1, that is, a
A circle with (1,2) as its center and 1 as its radius
So the value of Y, that is, the value of B is 2-1 "B" 2 + 1 "
So 1 B 3

Given the complex number Z = - 3-4i, then the imaginary part of 1 / Z is____
Reasoning process and what formula

Denominator materialization
1/z=1/(-3-4i)
=(-3+4i)/(-3+4i)(-3-4i)
=(-3+4i)/[(-3)^2-(-4i)^2]
=(-3+4i)/(9+16)
=(-3+4i)/25
So the imaginary part is 4 / 25

x²+（√5+√3）x+√15=0 （x²+x）（x²+x—2）=24 （x—√2）=√5x（2—√2x）
Solving quadratic equation of one variable

As a result, we're going to be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, and we will be in the 178, 178, and we will be in the 178, and we will be in the 178, 178, and we are going to be in the 178, 178, 178, and we are going to be in the 178, 178; and the 178; and we are going to be in the 178; and the 178; the 178; and the 178; and the 178; we are going to be in the 178; and the 178; 178; the 178; the 178; (178; (178; 178; 178; and the 178; (178; 178; 178; 178- 6) = 0 [(x + 1 / 2) &

（-x²+5x+4）+（5x-4=2x²）
Wrong
Yes (- X & # 178; + 5x + 4) + (5x-4 + 2x & # 178;)

To remove the brackets is - X & # 178; + 5x + 4 + 5x-4 + 2x & # 178;
Simplification is X & # 178; + 10x
Finally, X (x + 10)
（-x²+5x+4）+（5x-4+2x²）
＝x²＋10x＝x﹙x＋10﹚