So I assumed I had to do it in a way similar to the way I did.
This is probably stupid but when I do it that way, I end up with 2isin3θ since:
z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ isin(3\theta)- (cos(-3\theta)+ isin(-3\theta)= cos(3\theta)+ isin(3\theta)- cos(3\theta)+...
Could you explain this a little more?
I managed to solve it by drawing a rhombus with |z^3- z^{-3}|being the vertical length along the imaginary axis from the origin to the point z^3- z^{-3} (by definition) then drawing a perpendicular bisector (that is, parallel to the real axis) which...
Hi guys,
I've been trying to help a friend with something that I learned in class but I'm now finding it hard to solve myself. The problem goes as follows:
Use geometry to show that |z3-z-3| = 2sin3θ
For z=cisθ, 0<θ<∏/6
Now, I chose ∏/12 as my angle and plotted all this on an Argand diagram...
Ah, yes.
I realized before I fell asleep that the slope of the tangent at any point along that function with t on the x-axis is the velocity at that point in time (obviously).
Thanks for the help.
Hey guys,
I'm currently in year 11 (Australia) and my physics class has recently started projectile motion. I noticed in class that the elementary differential calculus I've been learning in math could be applied to the questions we are working on. I'd been itching to try it out and today...