Pls Help Roots of Complex numbers (1 Viewer)

[Lith_30]

I need help with part c

thankyou!

 

[tito981]

I need help with part c
View attachment 33028
thankyou!

note when dividing LHS by Z cubed you get:

z^3+z^-3

it is known that:

cis(nx)+cis(-nx)=2cos(nx)

Thus using this on the new LHS statement you gain, where n=3:

2cos(3 theta)

Now since you have to divide RHS by z^3, split up the division into dividing by z, 3 times. Now divide each of the brackets by z only thus:

RHS= (z+z^-1)(z-sqrt(3)+z^-1)(z+sqrt(3)+z^-1)

then collecting the applying the rule in the second line above you get:

(2cos(theta))(2cos(theta)-sqrt(3))(2cos(theta)+sqrt(3))

then taking 2 out of each of the brackets and dividing both RHS and LHS by 2 you get:

cos(3 theta) =4cos(theta)(cos(theta)-cos(pi/6))(cos(theta)+cos(5pi/6))

As required.

 

[Lith_30]

note when dividing LHS by Z cubed you get:

z^3+z^-3

it is known that:

cis(nx)+cis(-nx)=2cos(nx)

Thus using this on the new LHS statement you gain, where n=3:

2cos(3 theta)

Now since you have to divide RHS by z^3, split up the division into dividing by z, 3 times. Now divide each of the brackets by z only thus:

RHS= (z+z^-1)(z-sqrt(3)+z^-1)(z+sqrt(3)+z^-1)

then collecting the applying the rule in the second line above you get:

(2cos(theta))(2cos(theta)-sqrt(3))(2cos(theta)+sqrt(3))

then taking 2 out of each of the brackets and dividing both RHS and LHS by 2 you get:

cos(3 theta) =4cos(theta)(cos(theta)-cos(pi/6))(cos(theta)+cos(5pi/6))

As required.

Thankyou! Now it makes much more sense.

 

[5uckerberg]

What you see here is the application of De Moivre's Theorem, and note that adding a complex number to its inverse is the same as adding a complex number to its conjugate. As such it becomes noting that z is the complex number. It is shown that .

Note that the product of z and its conjugate becomes 1 as the radius is 1 if z is a general complex number

 

[CM_Tutor]

Note that the statement

What you see here is the application of De Moivre's Theorem, and note that adding a complex number to its inverse is the same as adding a complex number to its conjugate.

is true only if the complex number has a modulus of 1, for the reasons noted in

It is shown that .

Note that the product of z and its conjugate becomes 1 as the radius is 1.

For a general complex number , of any modulus, like , the result is true for all , but the result for does not generalise.

 

[5uckerberg]

Note that the statement

is true only if the complex number has a modulus of 1, for the reasons noted in

For a general complex number , of any modulus, like , the result is true for all , but the result for does not generalise.

Thank you for demystifying my work. Sometimes there needs to be a disclaimer so that people do not get misled.

 

[CM_Tutor]

Thank you for demystifying my work. Sometimes there needs to be a disclaimer so that people do not get misled.

No problem, I see part of my role as clarifying for the sake of other readers. I think it helps everyone to be confident that information is reliable, and I hope it encourages students to help each other, knowing that any mistakes will be corrected and issues clarified so that they can become part of the learning process.