Given that in the complex plane, the moving point Z corresponds to the complex z = x + Yi, then what graph is the set of points Z satisfying the equation Z ˉ 2 = 1

Given that in the complex plane, the moving point Z corresponds to the complex z = x + Yi, then what graph is the set of points Z satisfying the equation Z ˉ 2 = 1

∵lz-2l=l(x-2)+yil=1
∴(x-2)²+y²=1
So: the set of Z is a circle with (2,0) as the center and radius 1

Complex z = (1-m) + (1-m & sup2;) I, (m ∈ R), find the corresponding point of Z in the complex plane
A. Not in quadrant four
B. Not in the third quadrant
C. Only in the fourth quadrant
D. Not in the second quadrant

D
If M > 1, then 1-m

The complex number Z = (2 + I) m ^ 2-3 (1 + I) m-2 (1-I). When m is a real number, the corresponding point of Z in the complex plane is below the real axis

z=（2+i）m^2-3（1+i)m-2(1-i)
=2m²+m²i -3m-3mi -2+2i
=2m²-3m -2+（m² -3m+2）i
To make the corresponding point of Z in the complex plane below the real axis
That is, the imaginary part of Z is less than zero
That is M & sup2; - 3M + 2 < 0
The solution is 1 < m < 2

How to prove that | Z | = | Z ′| Z is a complex number and Z 'is a conjugate complex number

Let z = a + bi, a, B be real numbers, then Z '= a-bi
|Z | = root sign (a square + b square)
|Z '| = root sign (a square + (- b) square)
So | Z | = | Z '|