Given the complex number Z = 1-I, then Z2 − 2zz − 1=______ ．

Given the complex number Z = 1-I, then Z2 − 2zz − 1=______ ．

If z = 1-I, then Z2 − 2zz − 1 = (1 − I) 2 − 2 (1 − I) (1 − I) − 1 = − 2 − I = 2II · I = - 2I;

Solve the equation | Z ^ 2 | + (conjugate complex number of Z + Z) I = 3-I / 2 + 2 in the complex range

Let z = a + bi
Then a ^ 2 + B ^ 2 + 2ai = 7 / 2-I / 2
That is, a ^ 2 + B ^ 2 = 7 / 2
2a=-1/2
The solution is as follows
a=-1/4
B = 55 ^ 0.5/4 or - 55 ^ 0.5/4

Solving complex equation Z ^ 3-3iz - (3-I) = 0

The complex equation Z & # 179; - 3iz - (3-I) = 0 can be reduced to Z & # 179; + i-3iz-3 = 0
That is Z & # 179; - I & # 179; - 3I (Z + 1 / I) = 0
That is, (Z-I) (Z & # 178; + iz + I & # 178;) - 3I (Z-I) = 0
So (Z-I) (Z & # 178; + iz-1-3i) = 0
So z = I