(2x-1) + I = y - (3-y) iy, where y is a pure imaginary number and X is a real number, then x + y =?

(2x-1) + I = y - (3-y) iy, where y is a pure imaginary number and X is a real number, then x + y =?

This is equal to the plural
Let y = MI
It is concluded that Y - (3-y) iy = 3M + (M-M ^ 2) I = (2x-1) + I
2x-1=3m
(m-m^2)=1

If x is a pure imaginary number, y is a real number, and 2x-1 + I = y - {3-y} I find X and Y. it is better to have a process

2X-1+i=Y-{3-Y}i
Transfer: (- 1-y) + (3I Yi + 2x + I) = 0
So - 1-y = 0; 3I Yi + 2x + I = 0
5 I, y = - 1

Given that the complex number 1 + I is a root of the real coefficient equation x ^ 2 + ax + B = about X, then the value of 3A + 2b is
A.0
B.1
C.2
D.-2

The solution from 1 + I is the root of the real coefficient equation x ^ 2 + ax + B = 0 about X
That is, (1 + I) ^ 2 + a (1 + I) + B = 0
That is, 2I + A + AI + B = 0
That is, (a + b) + (2 + a) I = 0
That is, a + B = 0
a+2=0
That is, a = - 2, B = 2
So 3A + 2B
=-6+4
=-2
Announcement D