If Z + I is a real number and Z / (1-I) is a pure imaginary number, then Z + I is a complex number=

If Z + I is a real number and Z / (1-I) is a pure imaginary number, then Z + I is a complex number=

Let z = a + bi
Z + I = a + (B + 1) I is a real number,
Then B = - 1
therefore
z=a-i
z/(1-i)
=(a-i)(1+i)/(1-i)(1+i)
=(a+ai-i-i^2)/(1-i^2)
=(a+1 +(a-1)i)/2
Because it's a pure imaginary number
So a + 1 = 0
a=-1
So z = - 1-i
Given that one zero point of the function y = x ^ 3 + 2x ^ 2 + mx-6 is 2, find the other zeros of the function
f(x)=x^3+2x^2+MX-6
According to the meaning of the problem, f (2) = 8 + 8 + 2m-6 = 10 + 2m = 0, M = - 5
That is, f (x) = x ^ 3 + 2x ^ 2-5x-6
Because x = 2 is a root of F (x) = 0
So it can be factorized to (X-2) (x ^ 2 + 4x + 3) = 0
That is to say, x ^ 2 + 4x + 3 = 0 is factorized into (x + 1) (x + 3) = 0
The other two zeros are x = - 1 and x = - 3
Bring (2,0) in,
M=-5
The original formula is y = x ^ 3 + 2x ^ 2-5x-6
Divide by (X-2)
y=(x-2)(x^2+4x+3)
So the other zeros are - 1, - 3
You take x = 2 and find m, and then you factorize it.
In sequence an, A1 = 1, A1A2... An = n * 2, find A3 + A5
a1a2...a(n-1)=(n-1)*2 (n>=2）
Compare two formulas to get an = n ^ 2 / (n-1) ^ 2 (n > = 2)
Then A3 = 9 / 4, A5 = 25 / 16
So A3 + A5 = 61 / 16
Given that z is a complex number, (Z-1) / I is a real number, and Z / (1-I) is a pure imaginary number, find Z
let z=a+bi
(z-1)/i
=[(a-1)+bi]/i
=B - (A-1) I is a real number
=> a=1
z/(1-i)
=(1+bi)/(1-i)
=(1+bi)(1+i)/2
=(1 / 2) ((1-B) + (B + 1) I) is a pure imaginary number
=> b=1
z=1+i
Let z = a + bi
(z-1)/i=-i[(a-1)+bi]=-(a-1)i+b
(Z-1) / I is a real number
a-1=0
z/(1-i)
=(a+bi)/(1-i)
=[(a+bi)(1+i)]/2
=[(a-b)+(a+b)i]/2
The imaginary number is Z-1 / (Z-1)
a-b=0
a+b≠0
a=b=1
z=1+i
Let z = a + bi
If (Z-1) / I = (A-1 + bi) / I = [(A-1) I-B] / (- 1) is a real number, then A-1 = 0, a = 1
If Z / (1-I) = (a + bi) (1 + I) / (1 + 1) = (a-b + (a + b) I) / 2 is a pure imaginary number, then A-B = 0, that is, a = b = 1
Let z = 1 + I be z = a + bi
If (Z-1) / I = (A-1 + bi) / I = [(A-1) I-B] / (- 1) is a real number, then A-1 = 0, a = 1
Let f (x) = x ^ 2-2x + m, if f (x1) = f (x2), (x1, X2 are not equal to 0), then f is?
A2 B1 Cm Dm-1
Choose D
Because (x1 + x2) / 2 is the symmetry axis of the function, there is (x1 + x2) / 2 = 1
Substitute D
In the sequence {an}, we know A1 = 2, an + 1 = an / 3an + 1 (n ∈ n *), find the general term formula of A2, A3, A4 conjecture an, and prove it
a2=2/7;
a3=2/13;
a4=2/19;
an=2/(1+6(n-1));
Proof: mathematical induction is very simple
When n = 1, A1 = 2 is true,
2, suppose that when n = k, AK = 2 / (1 + 6 (k-1)); then when n = K + 1
AK + 1 = AK / (3AK + 1) = 2 / 6 + 1 + 6 (k-1) = 2 / 1 + 6K, that is, when n = K + 1, the proposition also holds,
From 1 and 2, we can get the proposition that an = 2 / (1 + 6 (n-1));
It's easier to use mathematical induction to make you guess first and then prove
It is proved that (1 / an) is AP
1/an+1=3+1/an
D=3
1/an=1/a1+(n-1)3
1/an=1/2+3n-3
1/an=6n-5/2
an=2/6n-5
Given that 1 + AI / 1 + I is a pure imaginary number (where I is an imaginary unit), then the real number a is equal to
(1＋ai)/(1＋i)
=[(1＋ai)(1－i)]/[(1＋i)(1－i)]
=(1/2)[(1＋a)＋(a－1)i]
Then: 1 + a = 0, a = - 1
Given the function f (x) = x ^ 2, G (x) = (1 / 2) ^ 2-m, if there exists x2 [0,2] for all X1 [- 1,2] such that f (x) is greater than or equal to G (x)
Find the range of M?
Is f (x1) greater than or equal to G (x2)
For any x1 ∈ [- 1,2], there exists x2 ∈ [0,2], such that f (x1) ≥ g (x2), then the maximum value of G (x) in [0,2] needs to be less than or equal to the minimum value of F (x) in [- 1,2] ∵ f (x) = x ^ 2 is monotonically decreasing in [- 1,0], and the minimum value of F (x) in the interval x1 ∈ [- 1,2] is f (0) = 0 ∵ g (x) = (1 / 2) ^ X -
∵ x1) > = g (x2) holds
The minimum of F (x) > the maximum of G (x)
The minimum value of F (x) - 1 / 8 when x = 3 / 4
When x = 0, G (x) = (1 / 2) ^ x-m, the maximum value of G (x) is 1-m
So 1-m = 9 / 8
The sequence an satisfies Sn = 3an-1 / 2, calculates A1, A2, A3 and A4, guesses the general term of an, and calculates the first n terms of an and Sn
A1 = S1 = 3a1-1 / 2, so A1 = 1 / 4, we can find the general formula of A2, A3, A4 in turn: SN = 3an-1 / 2, Sn + 1 = 3an + 1-1 / 2, ② - 1 → an + 1 = 3an + 1-3an, so an + 1 / an = 3 / 2, so {an} is the equal ratio sequence of common ratio q = 3 / 2, the general formula of {an} is: an = A1 * q ^ n-1 = 1 / 4 * 3 / 2 ^ n-1a1, Q
Given that a is a real number, if (1 + I) (2 + AI) is a pure imaginary number, then what is a equal to
A=2
Expansion = 2 + (2 + a) I + AI ^ 2
i^2=-1
ai^2=-a
（2-a）+（2+a）i
Because it is a pure imaginary number
2-a=0
A=2