Mtaule mgunda
Senior Member
- Jan 25, 2014
- 193
- 61
Solve For Z If CosZ=2 where z = X+iY
Solve For Z If CosZ=2 where z = X+iY
Cosz=2 is complex trigonometric equation soln as below:Solve For Z If CosZ=2 where z = X+iY
Solve For Z If CosZ=2 where z = X+iY
miss chagga wangu ndo manini hayo? Ma Cosmas sijui, sijui Z?Z= 0 ... ujasema tuonyeshe njia
Cosz=2 is complex trigonometric equation soln as below:
We have cos(z) = (e^(iz)+e^(-iz))/2 = 2.(Euler equation)
So e^(iz)+e^(-iz) = 4.
Multiply through by e^(iz) and we get
e^(2iz) + 1 = 4e^(iz).
So e^(2iz) - 4e^(iz) + 1 = 0. This is a quadratic in e^(iz)
Applying the quadratic formula:
e^(iz) = {4 +or- sqrt(16-4)}/2 = {4 +or- 2sqrt(3)}/2
= 2 +or- sqrt(3)
iz = ln(2 +or- sqrt(3))
z = -i*ln(2 +or- sqrt(3)) = i*ln{1/(2 +or- sqrt(3))}
= i*ln(2-sqrt(3)) or i*ln(2+sqrt(3))
hapo z haipo yaani hai-exist sababu ili upate z lazima upate cos inverse ya 2 wakati maxmum value ya cos ni moja
Hiyo hesabu utaitumia wapi kupata hela?
hakuanaga hesabu ya hivyomiss chagga wangu ndo manini hayo? Ma Cosmas sijui, sijui Z?
Hiyo hesabu utaitumia wapi kupata hela?
hakuanaga hesabu ya hivyo
Kaka Z =x +iy Cjaona Ku2mia Hyo Nazan Umekosa Co Kwel Hyo ,,, Pia Hyo Co Cosh N Cos
Cosz=2 is complex trigonometric equation soln as below:
We have cos(z) = (e^(iz)+e^(-iz))/2 = 2.(Euler equation)
So e^(iz)+e^(-iz) = 4.
Multiply through by e^(iz) and we get
e^(2iz) + 1 = 4e^(iz).
So e^(2iz) - 4e^(iz) + 1 = 0. This is a quadratic in e^(iz)
Applying the quadratic formula:
e^(iz) = {4 +or- sqrt(16-4)}/2 = {4 +or- 2sqrt(3)}/2
= 2 +or- sqrt(3)
iz = ln(2 +or- sqrt(3))
z = -i*ln(2 +or- sqrt(3)) = i*ln{1/(2 +or- sqrt(3))}
= i*ln(2-sqrt(3)) or i*ln(2+sqrt(3))
Keyser Söze;14715996 said:cos(z) = 2. We have cos(z) = (e^(iz)+e^(-iz))/2 = 2.
So e^(iz)+e^(-iz) = 4.
Multiply through by e^(iz) and we get
e^(2iz) + 1 = 4e^(iz).
So e^(2iz) - 4e^(iz) + 1 = 0. This is a quadratic in e^(iz)
Applying the quadratic formula:
e^(iz) = {4 +or- sqrt(16-4)}/2 = {4 +or- 2sqrt(3)}/2
= 2 +or- sqrt(3)
iz = ln(2 +or- sqrt(3))
z = -i*ln(2 +or- sqrt(3)) = i*ln{1/(2 +or- sqrt(3))}
= i*ln(2-sqrt(3)) or i*ln(2+sqrt(3))
Is it that one has copied the other, or, you have all copied from the same source!
Or else, the IDs belong to one person.