Complex Number (1 Viewer)

[Hello_World2]

I'm not too sure how to do parts a and b for Q8. Can anyone help, thanks !

 

[cossine]

View attachment 36623
I'm not too sure how to do parts a and b for Q8. Can anyone help, thanks !

Your image is rotated 90 degrees

 

[Hello_World2]

Your image is rotated 90 degrees

Yea ik but I'm not sure how to rotate, it doesn't give me a button for rotating

 

[Hello_World2]

Yea ik but I'm not sure how to rotate, it doesn't give me a button for rotating

Alright, I fixed it thanks for pointing it out

 

[jimmysmith560]

Would the following working help?

Part (a):



Part (b):

(i) From part (a):



(ii)

 

[Grinder]

Jimmy didn't know you were good at math

 

[jimmysmith560]

Jimmy didn't know you were good at math

Thank you Grinder, although I am not (I wish I was though haha).

However, what I seem to be good at is finding solutions to questions and I try to use this to help others.

 

[Hello_World2]

Would the following working help?

Part (a):

View attachment 36627

Part (b):

(i) From part (a):

View attachment 36628

(ii)

View attachment 36629

damn thanks a lot, you're a lifesaver! Just a quick question, how do we know that tan4θ is 1/√3 ?

 

[jimmysmith560]

damn thanks a lot, you're a lifesaver! Just a quick question, how do we know that tan4θ is 1/√3 ?

No worries! I sadly cannot help you much beyond this point, although perhaps @cossine could answer your question?

 

[cossine]

damn thanks a lot, you're a lifesaver! Just a quick question, how do we know that tan4θ is 1/√3 ?

You need to give the question a go. Often when we are try solve, prove etc we don't know how to approach the problem.

When I first look at the problem I realize coefficients were similar the polynomial, through inspection. After examining it a little bit more I realise tan(4*theta) = 1/sqrt(3) will result in the polynomial in part 8a).

So in other words by inspection.