When working in the real numbers, $\sin x$ has a nice geometric interpretation as the vertical coordinate that you arrive at after tracing an arc of $x$ units anticlockwise around the unit circle. Is there a similar way of defining $\sin x$ when $x$ is a complex number? I’m not particularly interested in the graph of $\sin x$; rather, I’m looking for a geometric interpretation of what $\sin x$ "means" in a way that is analogous to how we understand it using the unit circle.
Can complex sine be understood using the unit circle?
549 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 3 best solutions below
On
I've put together a three.js visualizer for cross sections and projections of the complex hypersurface $x^2+y^2=1$. There are a few artifacts on the projections due to the way I'm connecting points together.
https://trkern.github.io/ccircle.html
While it is true that $\sin^2(x)+\cos^2(x)=1$ even for complex $x$, does anyone have a good intuition for how the location of the point $(\sin(x),\cos(x))$ is related to $x$?
On
$\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Reals}{\Number{R}}\newcommand{\Cpx}{\Number{C}}$If $t = u + iv$ is complex (with $u$ and $v$ real), the mapping $\gamma(t) = (\cos t, \sin t)$ parametrizes the locus $x^{2} + y^{2} = 1$ in $\Cpx^{2}$ just as in the real setting, and is a covering map. (Analogously, $\exp$ is a covering map onto its image.) In this "complex" sense, $\sin z$ is still the second coordinate of the parametrization.
The image of $\gamma$ is the affine quadric: holomorphically a cylinder, i.e., the sphere with two points removed. In the complex projective plane, the two punctures lie on the line at infinity, and the quadric is a rational curve, holomorphically a sphere (but not a projective line).
The animation loop shows rotation of the real plane spanned by the imaginary parts of the complex Cartesian coordinates. In gory detail, $$ f_{\theta}(u, v) = (\cos u\cosh v, \sin u \cosh v, -\sin u\sinh v\cos\theta - \cos u \sinh v\sin\theta), $$ with $\theta$ running from $0$ to $2\pi$. The real points on the unit circle shown are fixed by these rotations. (A pair of planes in $\Reals^{4}$ can intersect in one point, and a rotation of one plane fixes the orthogonal plane.) The "circularly tilting" green ovals make visible that the surface has two planar ends. These intersect the (complex) line at infinity in one point each. The curves where the surface may appear to intersect itself, and the points where the surface may appear to pinch, are artifacts of the projection from $\Cpx^{2}$ to $\Reals^{3}$.

With $z=x+iy$ you can think of $\sin z$ as a vector, $$\sin z = (\sin x \cosh y,\cos x \sinh y).$$ Here's a plot of $w=\cosh y$ and $w=\sinh y$,
If you fix $y$ and let $\alpha=\cosh y$ and $\beta=\sinh y$ be constant then we obtain the vector $$\sin z=(\alpha\sin x,\beta\cos x).$$ That's an ellipse centred at the origin, excepting edge cases. It looks like for large $y$ you approach a circle of large radius. It's interesting to think of what happens to this ellipse as you travel left along the $y$ axis.
So to answer your question, the real part of $z$ is the angle of the arc traced anti-clockwise of an ellipse, the ellipse being defined by the imaginary part of $z$.