Given that the complex Z satisfies the square of complex Z + 2 times the modulus of conjugate complex z = 0, find the complex

Given that the complex Z satisfies the square of complex Z + 2 times the modulus of conjugate complex z = 0, find the complex

Let z = a + bi (a, B are real numbers)
Then (a + bi) &# 178; + 2 (a-bi) = 0
（a²-b²+2abi）+2（a-bi）=0
（a²+2a-b²）+（2ab-2b）i=0
therefore
1、a²+2a-b²=0
2、2ab-2b=0
If 2B (A-1) = 0, then B = 0 or a = 1
When B = 0, substituting into 1, we get a & # 178; + 2A = 0, a (a + 2) = 0, A1 = 0, A2 = - 2
When a = 1 is substituted into 1, we get 1 & # 178; + 2-B & # 178; = 0, B & # 178; = 3, B = ± √ 3
So Z1 = 0, Z2 = - 2, Z3 = 1 + √ 3I, Z4 = 1 - √ 3I
There are four solutions
Greatest common factor and least common multiple of 18, 36 and 54
18 and 108
A problem of derivation of complex numbers
If f (z) = Z * exp (a * cos (α) + b * sin (α)), Z is a complex number, α is the complex angle of Z, and a and B are constants, then f (z) is the derivative of Z, DF / DZ =?
First of all, don't talk nonsense if you think DF / DZ = exp (a * cos (α) + b * sin (α))!
If Z is a complex number, it should be in the form of Z (x, y) = x + I * y or Z (λ, α) = a * exp (λ + I * α) or Z = a * cos α + I * a * sin α,
Then, consider DZ / DX = 1; DZ / dy = I or DZ / D λ = = a * exp (λ + I * α); DZ / D α = a * exp (I) or DZ / D α = - A * sin α + I * a * cos α = I * (a * cos α + I * a * sin α) = I * Z
Should be wrong??
Let z = MCOs α + nsin α, then Z '= DZ / D α = NCOs α - MSIN α, then DF / DZ = (DF / D α) / (DZ / D α) and DF / D α = D [Ze ^ (ACOS α + bsin α)] / D α = D [(MCOs α + nsin α) e ^ (ACOS α + bsin α)] / D α = (NCOs α - MSIN α) e ^ (ACOS α + bsin α) + (MCOs
Let z = MCOs α + nsin α, then Z '= DZ / D α = NCOs α - MSIN α
Then, DF / DZ = (DF / D α) / (DZ / D α)
DF / D α = D [Ze ^ (ACOS α + bsin α)] / D α
＝d[(mcosα＋nsinα)e^(acosα+bsinα)]/dα
= (NCOs α - MSIN α) e ^ (ACOS α + bsin α) + (MCOs α + nsin α) [e
Let z = MCOs α + nsin α, then Z '= DZ / D α = NCOs α - MSIN α
Then, DF / DZ = (DF / D α) / (DZ / D α)
DF / D α = D [Ze ^ (ACOS α + bsin α)] / D α
＝d[(mcosα＋nsinα)e^(acosα+bsinα)]/dα
＝(ncosα－msinα)e^(acosα+bsinα)＋(mcosα＋nsinα)[e^(acosα+bsinα)](bcos
α－asinα)
And DZ / D α = NCOs α - MSIN α
∴df/dz＝(df/dα)/(dz/dα)＝e^(acosα+bsinα)＋(mcosα＋nsinα)[e^(acosα
+bsinα)](bcosα－asinα)/(ncosα－msinα)
＝e^(acosα+bsinα)＋z[e^(acosα+bsinα)](bcosα－asinα)/z’
＝e^(acosα+bsinα)＋(z/z’)(bcosα－asinα)[e^(acosα+bsinα)]
。 As far as this question is concerned, we can also do this:
df/dα＝d[ze^(acosα+bsinα)]/dα＝(dz/dα)e^(acosα+bsinα)＋z[e^(acosα
+bsinα)](bcosα－asinα)＝z’e^(acosα+bsinα)＋z[e^(acosα+bsinα)](bcosα
－asinα)
∴df/dz＝(df/dα)/(dz/dα)＝{z’e^(acosα+bsinα)＋z[e^(acosα+bsinα)]
(bcosα－asinα)}/z’
＝e^(acosα+bsinα)＋(z/z’)(bcosα－asinα)[e^(acosα+bsinα)]
Answer: put it away
df/dz=f/z+f·(-asinα+bcosα)dα/dz
Top 1L process~~
Generally speaking, it is not derivable.
＝e^(acosα+bsinα)＋(z/z’)(bcosα－asinα)[e^(acosα+bsinα)]
Eyes
What are the greatest common divisor and the least common multiple of 16 and 25?
What are the greatest common cause and least common multiple of 44 and 121, 5, 8, 10, 15, 18, 32, 24, 48, 72?
. greatest common divisor. Least common multiple
16 and 25.1.400
484
5、8、10.1.40
15、18、32.1.1440
24、48、72.24.144
How to find the derivative of e to the (1 + I) x power?
In this paper, we generalize the derivative of ^ (I) + X (g) + (E)) (x i) + (x i) + (x i) + (x E)) * (x i) + (x g) + (E)), and note that the derivative of ^ (I) + (x i) + (x i) + (x E) + (x i) + (x i) + (x i) + (x g) + (x E)) * (x i) + (x i) + (x i) + (x G) + (E)) (x i) + (x i) + (x i) + (x E) + (x i) + (x i) + (x i) + (x g) + (E)) (x i) + (x i) + (x I) + (E) + (x i) + (x i) + (x i) + (E)) (x i) + (x i) + (x i) + (x g * (e
y =e^(1+i)x
y' = (1+i) e^(1+i)x ( i is a constant )
Find the greatest common divisor and the least common multiple 40 and 25, 60 and 20, 11 and 33 of the following numbers
Find the greatest common divisor and the least common multiple of the following numbers
40 and 25, 60 and 20, 11 and 33
5 200
20 60
11 33
Write a function with C language to find the sum of two numbers
#include
int add(int a,int b)
{
return a+b;
}
main()
{
int a,b;
scanf("%d %d",&a,&b);
printf("a+b=%s",add(a,b));
}
The greatest common divisor and the least common multiple of 25 and 70
To use the method of factoring prime numbers, there are formulas
25=5x5
70=2x5x7
The greatest common divisor of 25 and 70 = 5, the least common multiple = 2x5x7 = 350
Let z = UV + Sint, u = e ', v = cost, and calculate the total derivative DZ / dt
What is u? What is e '?
The chain rule is used to find the total derivative of compound function
z=uv+sint
dz=vdu+udv+costdt;
u=e^t;
du=e^tdt;
v=cost
dv=-sintdt;
By substituting, we can get:
dz=v*e^tdt+u*(-sint)dt+costdt
=(ve^t+cost-usint)dt
=(cost*e^t+cost-e^tsint)dt.
dz/dt=(cost*e^t+cost-e^tsint).
Is e 'the derivative of E
The greatest common divisor of the sum of multiples is 18?
25=1×5×5
18=1×2×3×3
The greatest common divisor is 1, and the least common multiple is 2 × 3 × 3 × 5 × 5 = 450