Given the complex Z0 = 3 + 2I, the complex Z satisfies Z · Z0 = 3Z + Z0, then z = 1___ ．

Given the complex Z0 = 3 + 2I, the complex Z satisfies Z · Z0 = 3Z + Z0, then z = 1___ ．

Because Z0 = 3 + 2I, Z · Z0 = 3Z + Z0 is changed into Z · (3 + 2I) = 3Z + 3 + 2I, that is, 2zi = 3 + 2I х 2zi · I = 3I + 2I · iz = 1-32i, so the answer is: 1-32i

Let Z satisfy | z-z0 | = R, then what graph does it represent

|Z-z0 | denotes the distance between two points. So | z-z0 | = R denotes a circle with Z0 as the center and R as the radius

If the complex Z satisfies 3Z + | Z | = 17-9i, then=

Let z = a + bi, (a, B ∈ R)
Substituting 3Z + | Z | = 17-9i
That is 3A + 3bi + √ (A & # 178; + B & # 178;) = 17-9i
Ψ 3A + √ (A & # 178; + B & # 178;) = 17 and 3b = - 9
∴ b=-3
∴ 3a+√(a²+9)=17
That is √ (A & # 178; + 9) = 17-3a
∴ a²+9=289-102a+9a²
That is 8A & # 178; - 102a + 280 = 0
That is 4A & # 178; - 51A + 140 = 0
That is, (A-4) (4a-35) = 0
A = 4 or a = 35 / 4 (rounding, ∵ 17-3a ≥ 0)
∴ a=4,b=-3
That is Z = 4-3i