Is this a complex function?

Question

$f(\theta)=\cos(\theta)+i\sin(\theta)$, where $i^2=-1$ and $\theta \in [0,2\pi]$.

Is this function considered to be a complex function? If I want to differentiate it, should I verify the Cauchy-Riemann equations?

Thanks!

Answer 1

According to Wikipedia

A complex function is a function from complex numbers to complex numbers.

However, Wolfram says

A function whose range is in the complex numbers is said to be a complex function, or a complex-valued function.

So there is some ambiguity as to what the term "complex function" means. The domain of your function is the real numbers, while the range is the complex numbers. Thus, is you want be precise, you can call it a complex-valued function on the real numbers.

The Cauchy-Riemann equations apply to the meaning given by Wikipedia, that is, a complex-valued function on the complex numbers. So arguably they do not apply to the function you give. But there is an obvious extension to the domain of the real numbers. We can take $z = x+iy$ and $f(z) = u+iv$, and then $f(x+iy) = \cos(x) + i\sin(x)$, giving us $u_y = v_y=0$, which clearly leads to the CR equations not being satisfied. So $f$ is not differentiable with respect to $z$, but it is differentiable with respect to $x$: $\frac {df}{dx} = -\sin(x)+i \cos(x)$ (if we're taking this to be a function on the real numbers, this is a full derivative, but if we're taking it to be a function on the complex numbers, this should be a partial derivative).

Answer 2

This is a complex valued function of a real variable. It makes no sense to ask about the Cauchy-Riemann equations.

It is differentiable. Just go ahead and differentiate.

Geometrically, it wraps its domain around the unit circle in the complex plane. The derivative will be perpendicular to the value.

Answer 3

The domain is not an open subset of $\mathbb C$, and therefore the answer is negative.

It is differentiable because $\cos$ and $\sin$ are diffeentiable and $i$ is a constant. So, buy the usual rules of differentiation,$$f'(\theta)=-\sin(\theta)+i\cos(\theta)=i\bigl(\cos(\theta)+i\sin(\theta)\bigr).$$