### Re(cos(z)) on [-τ,τ]²

138
318
Views
Published on July 17, 2011

#### Description

The real component of cosine on the complex rectangle going from -τ (-2π) to τ (2π) on each side.

Notice that if you take a slice down Im(z) = 0, you get your normal cosine function, and slices Im(z) = τn | n ∈ ℕ get you cosh.

(To understand what is going on, recall that cos(x) = ½(e^(iθ) + e^(-iθ)), because the two exponentials, moving on the unit circle, have their imaginary components cancel but the real combine, giving twice the real component, cos(x). When you give a complex value, the imaginary component increases the radius of the circle one is moving on while decreasing the radius of the other, causing the to stop canceling. Since the radius chances exponentially, you get wider waves in addition to an imaginary component that you can't see here.)

#### Instructions

No instructions provided.
##### Tags
This Thing has no tags.
Report as inappropriate

Wait, is that right? Are you sure that cos(x) =
Â½(e^(iÎ¸) + e^(-iÎ¸))?

Well, I made switched from x to
Î¸ in typing, but other than that it is true. cos(Î¸) = Â½(e^(iÎ¸) + e^(-iÎ¸)). If you've ever heard that e^(iÏ) = -1, it's essentially the same phenomena: e^(iÎ¸) = cos(Î¸)+isin(Î¸).