[Year 12 Maths] Complex number - How am I supposed to approach part ii)
- Thread starter katsicum
- Start date
Hints: You can try writing cos(4theta) as cos(2(2theta)) or cos(2theta + 2theta) and then use a double angle or addition formula to proceed from there. Then you will have to change each cos(2theta) to something with theta since the right side of your equation has theta. If you get a sine squared, remember that sine squared = 1 - cosine squared as the right-hand side of your equation only has cosine and not sine.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,081
What did you get for part (i)? We can't do part (ii) without that answer, if indeed they are meant to be related.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
I for one have little or no idea about your year 12 course.
But the equation \(z^8=-1\) has eight complex roots.
Now the notation easy to learn \(z=r\exp(\theta {\bf i})=r\cos(\theta)+{\bf i} r\sin(\theta)\)
Thus we write \(-1=\exp(\pi{\bf i})=\cos(\pi)+{\bf i}\sin(\pi)\)
Then one of the eighth roots of \(-1\) is \(\zeta=\exp\left(\dfrac{\pi}{8}{\bf i}\right)\) that is called the principal eighth root.
No we spin that root around the unit circle to get seven more eighth roots.
Here is how we do it. Let say \(\rho=\exp\left(\dfrac{2\pi}{8}{\bf i}\right)=\exp\left(\dfrac{\pi}{4}{\bf i}\right)\)
Now having worked through all that notation here are all eight eighth roots of \(-1\):
\(\zeta\cdot\rho^k,~k=0,1,\cdots 7\).
But the equation \(z^8=-1\) has eight complex roots.
Now the notation easy to learn \(z=r\exp(\theta {\bf i})=r\cos(\theta)+{\bf i} r\sin(\theta)\)
Thus we write \(-1=\exp(\pi{\bf i})=\cos(\pi)+{\bf i}\sin(\pi)\)
Then one of the eighth roots of \(-1\) is \(\zeta=\exp\left(\dfrac{\pi}{8}{\bf i}\right)\) that is called the principal eighth root.
No we spin that root around the unit circle to get seven more eighth roots.
Here is how we do it. Let say \(\rho=\exp\left(\dfrac{2\pi}{8}{\bf i}\right)=\exp\left(\dfrac{\pi}{4}{\bf i}\right)\)
Now having worked through all that notation here are all eight eighth roots of \(-1\):
\(\zeta\cdot\rho^k,~k=0,1,\cdots 7\).