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Complex Numbers and Trigonometry


Date: 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/   
    
Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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