Complex Numbers and TrigonometryDate: 12/25/96 at 09:51:24 From: Thomas Glen Smith Subject: Why doesn't asin(sin(s)) = s for complex numbers? I calculate sin(3+4i) to be 3.85-27i, but when I compute arcsin(3.85-27i), I get .14-4i for an answer. Why don't I get 3+4i? I should think arcsin(sin(s)) should always produce s. It seems to work for arcsin(sin(.14-4i)). Date: 12/28/96 at 14:53:24 From: Doctor Pete Subject: Re: Why doesn't asin(sin(s)) = s for complex numbers? Hi, The underlying reason why this phenomenon occurs is because inverse trigonometric functions are not one-to-one when analytically continued to the complex plane. This also explains, for instance, why there are n n(th) roots of unity in the complex plane. To see what's going on in further detail, let's first look at a simpler example, the complex exponential. Euler's formula gives: Exp[i*t] = Cos[t]+i*Sin[t] where i is the imaginary unit Sqrt[-1]. Then the exponential of a general complex number in rectangular form z = a+i*b is: Exp[z] = Exp[a](Cos[b]+i*Sin[b]) From this, one can derive the inverse, the complex logarithm: Log[z] = Ln[Abs[z]]+i*(Arg[z]+2*Pi*k), k = ...-2,-1,0,1,2,... where Abs[z] is the magnitude of z, and Arg[z] is its *principal* argument, the angle it forms with the origin when plotted as a vector in the complex plane, restricted to some range, usually (-Pi, Pi]. Note that we have included a factor of 2*Pi*k. The reason for this is because there are infinitely many ways of measuring the argument; for instance, Arg[1+i] = Pi/4 but Arg[1+i] = {Pi/4, -7Pi/4, 9Pi/4, 17Pi/4, ... }. Therefore, while we see that the real part of Log[z] is single-valued, the imaginary part is multi-valued depending on what value of k we choose. Indeed, this is the dilemma at hand, for is there any particular reason why you should choose one k over another? We see then, that Exp[z], compared to Log[z], is a fairly simple function: for a given value of z, Exp[z] is unique, whereas Log[z] is actually an infinite set of values. But this is deceptive, for in reality, Exp[z] is just as complex (no pun intended) as Log[z]. A little thought shows that: Exp[a+i*b] = Exp[a+i*(b+2*Pi)] = Exp[a+i*(b+4*Pi)] = ... In other words, there are infinitely many values of z which have the same value of Exp[z]. So the full picture boils down to the fact that Exp[z] is a many-to-one function, and Log[z] is a one-to-many function. Furthermore, this strange behavior owes itself to the simple fact that Sin[x] and Cos[x] (for real x) are periodic functions with period 2*Pi. Now, it should not be too surprising, then, that we can use Euler's formula to obtain Sin[z] and Cos[z] for complex z. We have Exp[i*z] = Cos[z] + i*Sin[z] Exp[-i*z] = Cos[-z] + i*Sin[-z] = Cos[z] - i*Sin[z] Adding these two gives: 2*Cos[z] = Exp[i*z] + Exp[-i*z] Cos[z] = (Exp[i*z] + Exp[-i*z])/2 and subtracting gives: 2*i*Sin[z] = Exp[i*z] - Exp[-i*z] Sin[z] = (Exp[i*z] - Exp[-i*z])/(2*i) So the complex sine and cosine are expressible as sums of complex exponentials, and therefore are many-to-one functions. Therefore their inverses will behave like the complex logarithm in that they will be one-to-many. With this in mind, it is quite easy now to see why ArcSin[Sin[z]] is not necessarily z. For if z1 and z2 are unequal and such that Sin[z1] = Sin[z2], then ArcSin[Sin[z1]] is an infinite set which contains z1 and z2. Furthermore, ArcSin[Sin[z2]] is that same infinite set. Depending on which k-value you pick, you could even have ArcSin[Sin[z1]] = z2 and vice versa. However, if you knew what z1 was, then you could certainly pick ArcSin[Sin[z1]] = z1. But most calculators are not sophisticated enough to consider this fact, and instead simply choose the principal value (k=0) at all times. It may seem strange at first, but these phenomena are all rooted in the periodicity of the trigonometric functions. For example, ArcSin[Sin[x]] = ArcSin[Sin[x + 2*Pi]]. The ideas are the same, only with an added twist. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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