Vertical and horizontal nine grid, so that the horizontal, vertical, oblique addition is equal to a sum, can not be repeated Nine numbers from 1 to 9

Vertical and horizontal nine grid, so that the horizontal, vertical, oblique addition is equal to a sum, can not be repeated Nine numbers from 1 to 9

Pithy formula: wear nine shoes one, left three, right seven, two four for shoulder, six eight for foot
Both horizontal and vertical columns have three lattices, and the sum of the three numbers of each row, each column and two diagonals is equal, equal to 15
4 9 2
3 5 7
8 1 6
It's nine palace

How to make the addition of Y nine squares equal to nine and the numbers are not repeated

The sum of the lines in a nine palace grid equals 15, but not 9
Pithy formula: wear nine shoes one, left three, right seven, two four shoulders, six eight feet, five in the center
In fact, as long as you remember that "two or four shoulders, six or eight feet" is OK
Make the sum of the vertical, horizontal and oblique lines equal to 15
two hundred and ninety-four
seven hundred and fifty-three
six hundred and eighteen

Fill 1 ~ 7, 7 numbers into 9 squares, so that the sum of 3 numbers in each line and vertical line is equal to 12
Wrong, not 1 ~ 1 ~ 8 or 9 squares!

It's impossible
I didn't see your supplement
possible
1,8,3
6,4,2
5,,7

Reduction of imaginary number I ^ (7x + 2)
And I ^ (6x-1), I ^ (14x-11)

i^(6x-1)=i^6x/i=(i^2)^3/i=-i^(x-1)
i^(14x-11)=i^14x/i^11=(i^2)^7/-i=i^(x-1)

Simple imaginary number problem~~
The known equation x ^ 2 - (Tan θ + I) x - (I + 2) = 0
1. If the equation has real roots, find θ and two of them
2. It is proved that no matter what the value of θ is, the equation can not have pure imaginary roots

1.
-2 - tanθ x + x^2 - i (1 + x) = 0
If x is a real number, then
-2 - tanθ x + x^2 =0,
i (1 + x) =0
So x = - 1,
-2+tanθ+1=0,
tanθ=1,
θ = π / 4 + K π, K is an integer
So - 2 - x + x ^ 2 - I (1 + x) = 0,
(1 + x) (-2 - i + x)=0
X = - 1, or x = 2 + I
two
Suppose Tan θ is a real number, otherwise the problem is wrong
If the equation has pure imaginary roots, then
-2 - Tan θ x + x ^ 2 - I (1 + x)
Tan θ x is a pure imaginary number, or 0
-2 + x ^ 2 - I (1 + x) is a real number,
therefore
Tan θ x can only be 0, which contradicts that x is a pure imaginary number

imaginary number..
So I forgot, for example, a-2b + (a-b) I, which part of pure imaginary number is 0, which part is not 0? Non pure imaginary number? And so on,

Simple imaginary number calculation
(2I) / (1 + I) =? How to calculate

If the denominator is rationalized and multiplied by 1-I, the numerator becomes 2 + 2I and the denominator is 2, so the result is 1 + I

If the real part of the complex Z is positive and the imaginary part is 3, then in the complex plane, what graph should the point corresponding to the complex Z lie on?

It is a ray which is parallel to the real axis and points to the positive direction of the real axis with (0,3) as the starting point (excluding this point)

Given that the derivative of F (x) = 2x + 1x2 is f '(x), then f' (I) = (I is an imaginary unit) ()
A. -1-2iB. -2-2iC. -2+2iD. 2-2i

∵ f ′ (x) = 2x2 − 2x (2x + 1) X4 = − 2x2 − 2xx4, ∵ f ′ (I) = 2-2i, so D

If a, B ∈ R, I is an imaginary unit and (a + I) I = B + 52 − I, then a + B = ()
A. -2B. 0C. 1D. 2

∵ if a, B ∈ R, is an imaginary unit, and (a + I) I = B + 52 − I, ∵ AI + I2 = B + 5 (2 + I) (2 − I) (2 + I), it is reduced to - 1 + AI = B + 2 + I, ∵ 1 = B + 2A = 1, and the solution is a = 1b = − 3, ∵ a + B = - 2