Recent content by Quacknetar

Okay, I've just tried it and it's definitely easier and there's less things that you can do wrong. I wouldn't have thought of that myself, as ##Im(z)## doesn't feel like a "proper" function to me, but I see it can be useful if you know what you're doing. Thanks for the help, guys!

I used ##\sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}## How do you do it with ##Im(e^{ix})##?

Crap! I'm being stupid today! my previous expression should have been ##\frac{1}{2i}(\frac{1}{1-te^{ix}}-\frac{1}{1-te^{-ix}})## Which, if simplified, gives the same answer you got. Thanks, guys!

Damn, how did I not think of that? So the answer is ##\frac{2i}{2i-te^{ix}}-\frac{2i}{2i-te^{-ix}}##?

Homework Statement I have this exercise: Calculate ##\sum\limits_{k=0}^\infty t^{k}sin{(kx)}## Where x and t are real and t is between 0 and 1. Homework Equations ? The Attempt at a Solution The ratio test says that this sum does have a limit, and tk obviously converges, as t is between 0 and...