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During some experimentation with sines and cosines, its inverses, and complex numbers, I came across these results that I found quite interesting:

$ \sin ^ {-1} ( 2 ) \approx 1.57 - 1.32 i $

$ \sin ( 1 + i ) \approx 1.30 + 0.63 i $

Does this mean that there is such a thing as "imaginary" or "complex" angles, and if they do, what practical use do they serve?

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    $\begingroup$ the classical trigonometric functions defined on the real line with values in $[-1,1] $ admit holomorphic extensions with values in $\mathbb{C}$. The range is no longer $[-1,1]$, though, which kind of explains the result for $\sin^{-1}(2)$ I doubt that the output of the corresponding inverse functions still admits an interpretation as an angle. $\endgroup$ – user20266 Jan 3 '12 at 19:03
  • $\begingroup$ vqm.uni-graz.at/images/cfunc/sin3d.jpg $\endgroup$ – wim Jan 4 '12 at 5:07
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These values are relying on the generalization of the sine function as $$ \sin x = \dfrac{ e^{ i x } - e^{ - ix } }{2i}. $$ Clearly, when $x$ is complex it cannot be interpreted geometrically as an angle; however, generalized in this way, $\sin x$ becomes a holomorphic function, which is nice for a variety of other reasons. Practically, "imaginary angles" have some applications in physics. For instance, in optics, when a light ray hits a surface such as glass, Snell's law tells you the angle of the refracted beam, Fresnel's equations tell you the amplitudes of reflected and transmitted waves at an interface in terms of that angle. If the incidence angle is very oblique when traveling from glass into air, there will be no refracted beam: the phenomenon is called total internal reflection. However, if you try to solve for the angle using Snell's law, you will get an imaginary angle. Plugging this into the Fresnel equations gives you the 100% reflectance observed in practice, along with an exponentially decaying "beam" that travels a slight distance into the air. This is called the evanescent wave and is important for various applications in optics.

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    $\begingroup$ I was not aware of this generalization. This is so beautiful! $\endgroup$ – Avi Jan 4 '12 at 2:34
  • $\begingroup$ This also has application to acoustic waves incident on a boundary between two media. $\endgroup$ – orodbhen Apr 18 '17 at 15:55
  • $\begingroup$ Then maybe it has applications to all waves incident on a boundary between different media. Right? $\endgroup$ – afaolek Feb 12 '18 at 15:34
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See A geometric view of complex trigonometric functions by Richard Hammack in The College Mathematics Journal, May 2007; 38, 3 pages 210-217, for a much more thorough discussion of complex angle as hyperbolic angles.

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Yes, the sine function can be defined meaningfully on the whole complex plane as is explained here.

Whether or not you want to call a complex argument for the sine function an angle is a different question and, I suppose, a matter of taste.

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The Optics example can go farther:

Snell's Law states that $n_{0} \, \sin(\theta_{0})= n_{1} \, \sin(\theta_{1})$. For absorbing materials (or conductive like Au or Ag) the "$n_{1}$" is a complex number. This requires that $\theta_{1}$ also be complex.

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A fundamental equation of trigonometry is $x^2+y^2 = 1$, where $x$ is the "adjacent side" and $y$ the "opposite side".

If you experiment plot $f(x)$ out of the real domain - for example to $x=1.5$ you obtain $y$ imaginary - you will get an imaginary shape situated in a plane perpendicular to the plane $x,y$ and containing the $x$-axis. This shape is a hyperbola.

So you have two planes, one for the circle, and one for the hyperbola. The "$z$-axis" (imaginary) where the hyperbola is plotted correspond to the "$\sinh$" and $x$ is the "$\cosh$" once the $R = 1$. Note that the $\sinh$ is situated in a plane $90$ degrees of the $x,y$-plane. Observe that $$\sin iy = i \sinh y$$ is in accord with was explained above.

The geometric interpretation is easy.

It's valuable remember that the angle of a circumference could be measured by the double of the area of the sector. The hyperbolic angle could be measured by the double of area limited by the radius and the arc of hyperbola. See Wikipedia.

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  • $\begingroup$ The real reason for those "strange values" lies in Euler's formula $\mathrm{e}^{\mathrm{i} z} = \cos z + \mathrm{i} \sin z$, which can be solved to give $\cos z = \frac{\mathrm{e}^{\mathrm{i} z} + \mathrm{e}^{-\mathrm{i} z}}{2}$ and $\sin z = \frac{\mathrm{e}^{\mathrm{i} z} - \mathrm{e}^{-\mathrm{i} z}}{2 \mathrm{i}}$ $\endgroup$ – vonbrand Apr 29 '14 at 22:48
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Many areas in physics use what's called the Euler formula, which relates trigonometric functions to complex exponentials. This case, however does not actually admits complex angles: The arguement of the complex exponential is $e^{i\theta}$, which is interpted as either real trigonometric function $\cos(\theta)$ or $\sin(\theta)$.

On the other hand, a complex angle is a whole different story. I'll be giving now a qualitative physical explanation for why complex angles cannot be physical.

Let's assume that we have a certain physical system, say a pendulum - in which the energy depends on an angle $\theta$ by the equation $E=-mgL\cos(\theta)$. If we now allow the system to have hypothetical complex angle: $E=-mgL\cos(i\theta)$. Due to the identity $\cos(i\theta)=\cosh(\theta)$ we have an energy $E=-mgL\cosh(\theta)$. As we all know, this means that the energy is not bounded from below (as $\cosh(\theta)$ increase exponentially as $\theta$ tends to infinity) - and this is a serius break of the laws of physics. If it was true, then the system would tend toward the lowest energy state, and the conservation of energy demand it to release energy to the enviroment as the system minimizes it's energy. Thus creating infinite amount of energy out of the complex plane.

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  • $\begingroup$ Infinite energy on a complex plane? How can that happen? $\endgroup$ – Obinna Nwakwue May 10 '17 at 22:48
  • $\begingroup$ It can't happen. That's why complex angles aren't physical. $\endgroup$ – שחף רותם May 12 '17 at 11:14
  • $\begingroup$ So that means complex angles basically aren't real, right? $\endgroup$ – Obinna Nwakwue May 13 '17 at 16:28
  • $\begingroup$ Quite literally, yeah. $\endgroup$ – שחף רותם May 15 '17 at 19:56

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