If (a + I) (1-I) = 2, then the real number a =?

If (a + I) (1-I) = 2, then the real number a =?

solution
(a+i)(1-i)=2
Namely
a-ai+i+1=2
(a+1)+(1-a)i=2
∴1-a=0,a+1=2
∴a=1

Given that I is an imaginary unit and 1 / 2 of a + I is a pure imaginary number, what is the real number a equal to

Let (1 + I) / (a + I) = ki (where k is a real number not equal to 0)
Then: 1 + I = ki * (a + I)
(ka-1)i-(k+1)=0
So there are:
ka-1=0；
k+1=0
The solution is as follows
k=-1,a=-1
So a = - 1

A is a positive real number, I is an imaginary unit, | (a + I) / I | = 2, find a

A is a positive real number, I is an imaginary unit, | (a + I) / I | = 2, find a
|（a+i）/i|=|1-ai|=√（1+a^2）=2
a^2=3
a=±√3
A is a positive real number
So a = √ 3

Let z = 3 + 2I, then LM (the square of Z)=

Z^2=(3+2i)^2
=3^2+12i+4i^2
=9+12i-4
=5+12i