Solving complex equation: Z │ Z │ + AZ + I = 0 (a > = 0)
z=x+iy
(x+iy)(x^2+y^2)^(1/2)+a(x+iy)+i=0
There is x [√ (x ^ 2 + y ^ 2) + a] = 0
y√(x^2+y^2)+ay+1=0②
Get x = 0 or x = y = a = 0 from ① (not satisfied with ②, rounding off)
So y │ y │ + ay + 1 = 0
When y > = 0, there is no solution
When y
RELATED INFORMATIONS
- 1. The complex Z satisfying the equation Z2 + | Z | = 0 has () A. 1 B. 2 C. 3 d. countless
- 2. There are several roots of the equation 1-z Λ 4 = 0 in the complex range
- 3. Complex z = (square of a-2a + 3) - (square of A-A + 1 / 2) I (a belongs to R) in which quadrant is the corresponding point in the complex plane
- 4. Given the third quadrant of the point corresponding to the complex number Z = m (1 + I) - M 2 (4 + I) - 6I in the complex plane, the range of real number m is obtained
- 5. If the point corresponding to the complex z = (m-1) + (M & # 178; - 4) I (m ∈ R) in the complex plane is in the fourth quadrant, then M is in the range
- 6. Given | x + Yi | = 2 (x, y ∈ R), in the complex plane, find the set of points representing complex x + Yi Seek detailed process
- 7. If the complex z = (m-1) / 3 - (m-2) I (m ∈ R), its corresponding point on the complex plane is Z, then the shortest distance from point (1,2) on the complex plane to point Z is? The answer is = 2 √ 10 / 5
- 8. If the complex number Z is on the circle | Z | = 2, then it is proved that: | 1 / (Z ^ 4-4z ^ 2 + 3)|
- 9. If the conjugate complex number of Z + Z = 4 and the conjugate complex number of Z * z = 8, then the conjugate complex number of Z / Z is equal to? Detailed point answer, thank you
- 10. Given the complex number Z = 2 − 2i1 + I, then the conjugate complex number of Z is equal to () A. 2iB. -2iC. iD. -i
- 11. In the complex set C, the solution of the equation | x | + x = 1 + 3I is
- 12. In the complex range, the solution of the equation x ^ 2 + x-1-3i = 0 is?
- 13. Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = result is 2. How to do it? Another problem is in the complex range, the number of roots of the equation Z ^ 3 + | Z | = 0? Let f (z) = x ^ 2 + I * y ^ 2, then f '(1 + I) = the result is 2. How do you do that? Another problem is that in the complex range, how many roots of the equation Z ^ 3 + | Z | = 0? The answer to the second question is four roots. I'm very tangled
- 14. Solve the complex equation Z ^ 3 = I,
- 15. The complex Z satisfying the equation Z2 + | Z | = 0 has () A. 1 B. 2 C. 3 d. countless
- 16. On the equation Z2 - (a + I) Z - (I + 2) = 0 (a ∈ R) of complex number Z, (1) if the equation has a real number solution, find the value of a; (2) prove with the method of counter proof that for any real number a, the original equation cannot have pure imaginary root
- 17. To solve the equation Z ^ 2 + 2Z + 1-I = 0 in the complex range
- 18. Given that the complex Z satisfies the equation z2-2z + 3 = 0, then | Z|=______ .
- 19. Given that the complex Z satisfies 3Z + | Z | = 17-9i, find the complex Z
- 20. If complex x0 = 3 + 2I and Z * Z0 = 3Z + Z0, then complex Z=