How to calculate (1 + I) ^ 20 - (1-I) ^ 20,

How to calculate (1 + I) ^ 20 - (1-I) ^ 20,

(1+i)^20-(1-i)^20
=((1+i)^2)^10-((1-i)^2)^10
=(1+i^2+2i)^10-(1+i^2-2i)^10
=(2i)^10-(-2i)^10
=0
=(2i)^10-(-2i)^10=0
Trigonometric function on a problem solving!
If a > 0, b > 0, (a + 1) (B + 1) = 2, we prove that arctana + arctanb = π / 4
(a+1)(b+1)=2
ab+a+b+1=2
a+b=1-ab
tan(arctana)=a,tan(arctanb)=b
So tan (arctana + arctanb) = [Tan (arctana) + Tan (arctanb)] / [1-tan (arctana) Tan (arctanb)]
=(a+b)/(1-ab)
=1
-π/2
Let arctana = α, arctanb = β
∴a=tanα b=tanβ
∴(tanα＋1)(tanβ＋1)=2
The expansion result is tan α Tan β + Tan α + Tan β = 1
∴1-tanαtanβ=tanα+tanβ
And Tan (α + β) = (Tan α + Tan β) / (1-tan α, Tan β) = 1
∴α+β=π/4
That is arctana + arctanb = π / 4
More and more sharp
The past tense, present progressive tense of English verbs in Grade Seven
What are the features of the past tense, present progressive tense and other sentence forms of English verbs in grade seven
Ago, yesterday, last night and so on are often used in the past tense
Now, right now often appears in the progressive tense
How to get the process of complex number calculation (1 / - I) + (3 / 1 + I)
Solve several trigonometric function problems, urgent!
1. The minimum positive period of the function f (x) = sin2x-cos2x is ()
2. A is the fourth quadrant angle, cosa = 12 / 13, Sina = ()
3. In the triangle ABC, if Tana = 1 / 3, C = 150 ° and BC = 1, then AB =?
The reduction of problem 1 is f (x) = root sign 2Sin (2x - π / 4)
Period T = 2 π / 2 = π
Question 2 because sin is negative in the fourth quadrant, so Sina = root 1-cos ^ 2A = - 5 / 13
Tana = Sina / cosa = 1 / 3 a belongs to (0,2 π)
Sin ^ 2A + cos ^ 2 = 1. From this equation system, sin a = root 10 / 10
From the sine theorem, BC / Sina = AB / sinc is obtained
And sinc = 1 / 2, BC = 1, ab = radical 10 / 2
I don't know if the result is right, but the process is right
The answer period of the first question is π
Question 2 = - 13
1）π
2）-5/13
3) 1 / 2 * root 10
1. F (x) = 2 * sin (2x - π / 4) under the root sign, so t = 2 π / 2 = π
2. A is the four quadrant angle, so sin is negative. According to the square of sina + the square of cosa = 1, Sina = - 5 / 13
3. According to Tana = 1 / 3, it becomes Sina / cosa = 1 / 3. According to the square of sina + cosa = 1, we can find out 1 / 10 under the root sign of sina = 1. According to the sine theorem, Sina / BC = sinc / AB, we can find out 10 under the root sign of AB = 0.5
1. F (x) = 2 * sin (2x - π / 4) under the root sign, so t = 2 π / 2 = π
2. A is the four quadrant angle, so sin is negative. According to the square of sina + the square of cosa = 1, Sina = - 5 / 13
3. According to Tana = 1 / 3, it becomes Sina / cosa = 1 / 3. According to the square of sina + the square of cosa = 1, we can get the 1 / 10 under the root sign of sina = 1. According to the sine theorem, Sina / BC = sinc / AB, we can get the 10% under the root sign of AB = 0.5
What kinds of tenses does English have? The present continuous tense, the past tense, and so on?
The present tense
2. General past tense
3. General future
Past future tense
5. Present continuous tense
Past continuous tense
7. Future continuous tense
In the future
9. The present perfect tense
Past perfect tense
11. Future perfect tense
Past future perfect tense
The present perfect continuous tense
Past perfect continuous tense
15. Future perfect continuous tense
Past future perfect continuous tense
There are only nine tenses in common use
There are 16 tenses
According to time, tenses can be divided into four types: past tense, present tense, future tense and past future tense;
According to the action, it can be divided into four kinds: General tense, progressive tense, complete tense and complete progressive tense
Therefore, there are 16 kinds of combination, which can be used according to the specific language environment to determine which kind of time is and which kind of action belongs to
1. I did; 2;
The simple present tense (I do);
I will do;
How to solve the complex number with numerator and denominator
The numerator denominator is multiplied by the conjugate complex number of the denominator at the same time (the complex numbers whose real parts are equal and imaginary parts are opposite to each other are conjugate complex numbers), and then the denominator becomes a real number. For example: (1 - I) / (1 + I) = [(1 - I) (1 - I)] / [(1 + I) (1 - I)] = - I
Let's give you an example. Let's talk about it~~
First rationalize the denominator, then simplify it
How to rationalize the denominator
The problem of solving trigonometric function
Let f (x) = sin (ω x + φ), where ω > 0, | φ|
(1) The original formula is cos (π / 4) cos φ – sin (π / 4) sin φ = 0
=cos(π/4+φ)=0
π/4+φ=π/2
φ=π/4
(2) The distance between two axes of symmetry = π / 3
The period of this function is t = 2 π / 3
T=2π/ω ω=3
The result is f (x) = sin (3x + π / 4)
Verb change ing, past tense method, noun change plural, what change y to I go to e double write those at the end
The verb ends with E, goes to e and adds ing, such as dance dancing; if the ending is auxiliary, double write the last consonant and adds ing, such as dance dancing
The verb "hop hopping" changes to the past tense, ending with the consonant "Y", changing "Y" into "I" and "Ed", such as "study studied"; if the ending is the letter "e", add "d" directly, such as "dance danced"; if the ending is the auxiliary element, add "d" in double writing, such as "hop hopped"
The singular of the noun changes to the plural, ending with the consonant y, changing y into I and adding es; if there is e at the end, directly adding s; if the end is a complement, double writing and adding es
Complex multiplication
（cosX-i）（sinX+i）=?
0.5 * sin2x-1, right?
You are wrong!
(cosx-i) (SiNx + I) expansion is not equal to cosxsinx-i * I, and - I * SiNx + I * cosx! You missed two terms
It should be like this:
（cosx-i）（sinx+i）=cosxsinx-i*sinx+i*cosx-i*i
=0.5*sin2x-1-（sinx-cosx)i
cosxsinx-i*sinx+i*cosx+1
Don't think about it. That's the answer
There is almost no difference between the four operations of complex numbers and real numbers. Similarly, if you expand polynomials, just remember I * I = - 1.