Why is it 3 π / 4 to find the modulus and the principal value of the radiation angle of the complex z = - 1 + I

Why is it 3 π / 4 to find the modulus and the principal value of the radiation angle of the complex z = - 1 + I

According to the definition of radiation angle, the starting edge is in the positive direction of X axis, and naturally the principal value of radiation angle is 3 π / 4
Draw a picture and come out
The complete set u = R, the set a = {x | x > 0}, B = {x | x ^ 2 ≤ 4}, find CUA ∩ B, a ∪ cub
Today's homework, to say shame, I really don't know how to ask for the complete works of a and B
Please tell me how to solve the problem. I already know the information of 0 < x ≤ 2. I don't know how to think about it. If I know, let me know,
Complete set u = R, a = {x | x > 0}, B = {x | x ^ 2 ≤ 4} = {x | - 2 ≤ x ≤ 2}
So CUA = {x | x ≤ 0} cub = {x | x ＜ - 2 or X ＞ 2}
（CuA）∩B=｛x|-2≤x≤0｝
A ∪ (cub) = {x | x ＜ - 2 or X ＞ 0}
A∩B=｛x|0＜x≤2}
Cu (a ∩ b) = {x | x ≤ 0 or x > 2}
Note:
I don't know which CUA ∩ B means. Let's have a look. I wrote both
The complete set is r, of course. CUA means the part left after r removes a
B={x|x^2≤4}，
= {x| -2≤x≤2}
CuB={x| x>2 or x0}
CuA = {x| x ≤0}
CuA∩B = {x| x ≤0} ∩ {x| -2≤x≤2}
= {x| -2≤x≤0} #
A∪CuB = {x|x>0} ∪ ={x| x>2 or x0 or x
Given that the sum of the first n terms of the sequence {an} is Sn, and Sn = 6N an, then the general term formula of the sequence {an} is an=______ .
When n = 1, A1 = S1 = 6 × 1-a12a1 = 6A1 = 3N ≥ 2, an = SN-S (n-1) = 6n-an - [6 (n-1) - A (n-1)] 2An = a (n-1) + 62an-12 = a (n-1) - 6 (an-6) / [a (n-1) - 6] = 1 / 2, is the fixed value a1-6 = 3-6 = - 3, and the sequence {an-6} is an equal ratio sequence with - 3 as the first term and 1 / 2 as the common ratio, an-6 = (- 3) · (1 / 2) ^ (n
On the principal value of complex argument
The principal value of the complex z = - 3 (cos15 + isin15)
Is 15 an angle or a radian, an angle
z=-3(cos15°+icos15°)
=3(-cos15°-icos15°)
=3cos15°(-1-i)
=(3√2cos15°)(cos225°+isin225°)
The main value of radiation angle is 225 degrees
Let {a = ゝ u} be set {x} ゝ 3
Solution of CUA ∩ cub
=Cu（A∪B）
From the set a = {x | - 3 ＜ x ＜ 3}, B = {x | - 0 ＜ x ＜ 5}
Then a ∪ B = {x | - 3 ＜ x ＜ 5}
be
CuA∩CuB
=Cu（A∪B）
={x | - x ≥ 5 or X ≤ - 3}
Given the first few terms of the sequence (n-1 power an of 2) and Sn = 9-6n, then the general term formula of the sequence an is?
When n = 1, A1 = S1 = 9-6 = 3
When n is greater than 2, let BN = 2 ^ (n-1) an, so SN-S (n-1) = BN = - 6
So an = - 6 / 2 ^ (n-1)
Two kinds of situations, write it by yourself, I'm in trouble
What is the principal value of the argument of complex z = 1-I
The interval of principal value of radiation angle is (- π, π]
Z = 1-I, Z is in the fourth quadrant in the complex coordinate system
The main value of radiation angle is - π / 4
Given the complete set u = R, set a = {x | - 2 ≤ X
A:
Set a, - 2
If u = R, a = {x | - 2 ≤ x ≤ 3}, B = {x | X4}, then a ∩ (C ∪ b) is equal to ()
U=R
B = {x | x ＜ - 1 or X ＞ 4}
So cub = {x | - 1 ≤ x ≤ 4}
A={x|-2≤x≤3}
So a ∩ (cub) = {x | - 1 ≤ x ≤ 3}
Given the general term formula an = 2n-11 of sequence an, if BN = the absolute value of an (n belongs to n), find the sum of the first n terms of sequence BN
It is not obvious that substituting s (n) into S (1) is wrong
Obvious a (n) arithmetic sequence: so s (a (n)) = (a (1) + a (n)) * n / 2 = (N-10) * n
The following categories are discussed:
When n = 6, s (b (n)) = ABS (s (a (n))) + s (a (n))
Where ABS (x) is the absolute sign
Hello!!!
B6=1
So,
S（Bn）=1+3+5+7+9+（N-6）×N
S（Bn）=N^2-6N+25
Ask me if you don't understand.
thank you!!!
an=11-2n>0，n
What is the principal value of the argument of a complex number
What is the principal value of the argument of complex number-1-radical 3I? What are the specific steps to solve the problem?
It's - 1 minus the root sign 3 times I, not 3I under the root sign
z=-1-√3i=-2(1/2+√3/2i)=-2(cos60+sin60i),
=2[cos(180+60)+sin(180+60)i]
=2(cos4∏/3+sin4∏/3i).
The principal value of the argument of complex number-1-radical 3I is 4 Π / 3,