Atakayeweza Swali Hili La "complex Number'' Adv Math

[Mtaule mgunda]

Solve For Z If CosZ=2 where z = X+iY

 

[Preta]

Woooo........

 

[real G]

Solve For Z If CosZ=2 where z = X+iY

hapo z haipo yaani hai-exist sababu ili upate z lazima upate cos inverse ya 2 wakati maxmum value ya cos ni moja

 

[miss chagga]

Z= 0 ... ujasema tuonyeshe njia

 

[simplemind]

Solve For Z If CosZ=2 where z = X+iY

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))

 

[Mbilikimo Mfupi]

Kitambo sana mambo haya jaman daaaaa

 

[Nje ya Mada]

Solve For Z If CosZ=2 where z = X+iY

Z= 0 ... ujasema tuonyeshe njia

miss chagga wangu ndo manini hayo? Ma Cosmas sijui, sijui Z?

 

[Keyser Söze]

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))

 

[Mtaule mgunda]

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))

Kaka Z =x +iy Cjaona Ku2mia Hyo Nazan Umekosa Co Kwel Hyo ,,, Pia Hyo Co Cosh N Cos

 

[KakaJambazi]

Hiyo hesabu utaitumia wapi kupata hela?

 

[Mtaule mgunda]

hapo z haipo yaani hai-exist sababu ili upate z lazima upate cos inverse ya 2 wakati maxmum value ya cos ni moja

Hamna Kaka Hyo Unaweka Kwa Mtndo Wa Hyperbolic Ambapo Kuna Cosh Inverxe Ya 2

 

[MO11]

Hiyo hesabu utaitumia wapi kupata hela?

sijui
huu ujinga ndio maana hesabu zilinishinda

mtu anaweza akatunga swali halafu asilijue jibu ndio upuuzi wa hesabu

 

[miss chagga]

miss chagga wangu ndo manini hayo? Ma Cosmas sijui, sijui Z?

hakuanaga hesabu ya hivyo

 

[Don255]

Hiyo hesabu utaitumia wapi kupata hela?

Haha haha haaaa

 

[Ng'wanambula]

Yeah.... miaka 20 iliyopita nikiwa A-Level. So long!

 

[Raimundo]

Atakayeweza unampa nini?

 

[Nje ya Mada]

hakuanaga hesabu ya hivyo

Daaah.. Nilitaka nishangae unajipendekezaje kwenye hesabu zisizo na pesa!?

 

[simplemind]

Kaka Z =x +iy Cjaona Ku2mia Hyo Nazan Umekosa Co Kwel Hyo ,,, Pia Hyo Co Cosh N Cos

Unaelewa eular formula? Z=x+iy=re^IQ
Where r=sqrt (x^2+y^2) Q is an angle whose tangent is x/y.

 

[Raimundo]

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))

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.

 

[simplemind]

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.

La muhimu kuzingatia ni method. Trick ipo kutumia eular formula na solution of a quadratic equation. Anybody solving equation will use the same method. Copy and paste was merely to save me trouble of typing the solution.