Root I. 1 2. The cubic root of I Radical of 3.3 + 7I The third power of 4.1-i

Root I. 1 2. The cubic root of I Radical of 3.3 + 7I The third power of 4.1-i

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Do it according to the geometric meaning of complex root
Root I = e (I * pi / 4) = root 2 / 2 + I * root 2 / 2,
The cubic root of I = e ^ (I * pi / 6) = root 3 / 2 + I / 2,
Similarly, for the open radical of 3 + 7I, its module = root 58, its argument = arctan [7 / 3], and the open radical of 3 + 7I = root (root 58) (COS [arctan [7 / 3] / 2] + I * sin [arctan [7 / 3] / 2])
The open cubic power of 1-I, the module expansion of 1-I
Do it according to the geometric meaning of complex root
Root I = e (I * pi / 4) = root 2 / 2 + I * root 2 / 2,
The cubic root of I = e ^ (I * pi / 6) = root 3 / 2 + I / 2,
Similarly, for the open radical of 3 + 7I, its module = root 58, its argument = arctan [7 / 3], and the open radical of 3 + 7I = root (root 58) (COS [arctan [7 / 3] / 2] + I * sin [arctan [7 / 3] / 2])
The open cubic power of 1-I, the module of 1-I = root 2, the argument = - pi / 4,
Therefore, the open cubic power of 1-I = 2 ^ (1 / 6) * (COS (- pi / 12) + I * sin [- pi / 12]) is retracted
First, we write all these complex numbers in the form of R · e ^ (I θ) (R is the module length and θ is the auxiliary angle), and then we do it according to the geometric meaning of complex square root
But note that in the plural sense, there are two square roots and three cubic roots. If it is an open root sign, it is any one of these.
1. Because I = e ^ (I ·Π / 2), the root sign (I) = e ^ (I ·Π / 4) or e ^ (I · 3 Π / 4)
That is, the root sign (I) = root (2) / 2 + I * root (2) / 2 or - root (2) / 2-I * root
First, we write all these complex numbers in the form of R · e ^ (I θ) (R is the module length and θ is the auxiliary angle), and then we do it according to the geometric meaning of complex square root
But note that in the plural sense, there are two square roots and three cubic roots. If it is an open root sign, it is any one of these.
1. Because I = e ^ (I ·Π / 2), the root sign (I) = e ^ (I ·Π / 4) or e ^ (I · 3 Π / 4)
That is, the root sign (I) = root (2) / 2 + I * root (2) / 2 or - root (2) / 2-I * root (2) / 2
2. Because I = e ^ (I ·Π / 2), the cubic root (I) = e ^ (I ·Π / 6) or e ^ (I ·Π / 2) or e ^ (I · 5 Π / 6)
That is, cubic root (I) = root (3) / 2 + I / 2 or - I or root (3) / 2-I / 2
3. 3 + 7I = root (58) e ^ (I · arctan (7 / 3))
Therefore, root (3 + 7I) = + or - fourth root (58) e ^ (I · arctan (7 / 3) / 2)
From the half angle formula sin (arctan (7 / 3) / 2)=
4. 1-I = root (2) e ^ (i.7 Π / 4), so the third root (1-I) = the fourth root (2) e ^ (i.7 Π / 8) or the fourth root (2) e ^ (i.15 Π / 8)
i=1*e^(PI/2*i)
The root sign is to divide pi / 2 by several times
Sorry, I can only answer one question
2。 The cubic root of I is equal to I times the root sign I
1.（√2）/2+(（√2）/2)i
2.-i
Making sentences with English phrases
Make two phrases for each
1.be angry with
2.a balanced diet
3.ask for advice
1. You should not be angry with him. I'm angry with people who is always late
Just look in the dictionary
Is something everything singular or plural? Is the verb be followed by is or are? Is it the same for somewhere and everywhere?
Is someone and everyone singular or plural? Are someone and everyone the same as someone and everyone
They or it
These are indefinite pronouns. The verbs after them are singular. The verb be should be is
Ask a few plural questions
1) If Z 2 + 2|z | = a (a ≥ 0), find the complex Z
2) If the complex Z satisfies | Z | = | Z + 2 + 2I |, then the minimum value of | Z-1 + 2I |, is________
3) If the complex Z satisfies 2|z-3-3i | - |z | = 0, then the value range of |z | is__________
Please understand the friend to help explain... Hope to have a process
1) If Z | 2 + 2 | Z | = a (a ≥ 0), the complex Z | Z | and a are both real numbers, so Z is a real number, z = A / 42) if the complex Z satisfies | Z | = | Z + 2 + 2I |, then the minimum value of | Z-1 + 2I | is________ |Z-0 | = | Z - (- 2-2i) | denotes Z to 0 and - 2-2i, that is, Z's trajectory is the middle vertical line of 0 and - 2-2i | Z - (1-2i) | denotes Z to 1-2i
30 English phrases and sentences
be poor at She is poor at swimmingtake care of She takes care of sick animalsbeware of Beware of sharks!at least At least you have a home.aware of He is aware of her presence.come In contact with ...
AwithB…… Is the verb be singular or plural
Kate and Lucy go to school everyday.
Kate with Lucy goes to school everyday.
With and the words after it do not affect the singular and plural of the subject
The verb be should be consistent with the subject, the subject is a, so we should look at the singular and plural of a: if a is an uncountable noun, use the singular; if a is a countable noun, use the plural
It is the same as the simple complex form of A
singular
Ask a few questions about the plural
First: when z = (i-1) / √ 2, the value of Z ^ 100 + Z ^ 50 + 1 is equal to?
Second: complex z = 1 + cos α + I * sin α (π)
Question 1: Z ^ 2 = - I
z^100=(z^4)^25=1
z^50=(z^4)^12*z^2=-1
z^100+z^50+1=1
Question 2: π
Make sentences with English phrases. Simple ones will do
A bit of
A few of
A kind of
a little of
a lot of
a pair of
a number of
a piece of
a set of
It's easy to write sentences with this
There is a bit of ink here. A little of them know it. A few of them know it
Countable noun plural, be verb use are; uncountable noun such as people, be verb use what?
First of all, people is not an uncountable noun
Second, when people means "person", it is usually a collective noun, so the predicate verb is usually plural
Of course, there are several people, such as ten people
At least ten people were killed in the crash
People is not uncountable, it is a collective noun, itself is plural. Use are
People as "people" is not countable, use are; but when people as "nation" is countable, use is or are. There are 36 people in China
People itself is a collective noun
When people refers to "nation", it can be regarded as a countable noun. Two peoples.
A question about the plural,
Given that 1-I is the root of the real coefficient equation x ^ 4-3x ^ 2-2ax + B = 0, find the value of a and B
Substituting 1-I into the equation, we get
(1-i)^4-3(1-i)^2-2a(1-i)+b=0
-4+6i-2a(1-i)+b=0
(6+2a)i-4-2a+b=0
6+2a=0,-4-2a+b=0
a=-3,b=-2
The conjugate complex number 1 + I of the complex number is also the solution of the secondary equation. This is the conclusion of a theorem. If the two solutions are replaced by the original equation, then a and B can be obtained. Let's do it by ourselves
Because the square of 1-I is - 2I, its fourth power is - 4. Substituting 1-I into the equation, we get - 4 + 6i-2a (1-I) + B = 0
So (6 + 2a) I + b-4-2a = 0
6+2a=0
b-4-2a=0
a=-3
b=-2