Some mathematics exercises
1. Raising a number to an imaginary power doesn’t immediately mean anything, so let’s accept as a
hypothesis that eix = cos(x) + i sin(x) . Then use this to evaluate both sides of ei(x+y) = eix eiy and interpret
the results.
2. Calculate the first and second derivatives (with respect to x) of both sides of eix = cos(x) + i sin(x) and
interpret the results.
3. One place where the relationship exp(ix) = cos(x) + I sin(x) turns out to be very useful is in the
discussion of periodic motion. There we are usually more interested in the frequency of the motion
than in the phase, so that in the relationship
x = A cos(w t + a)
it doesn’t matter too much what “a” is; yet we need to consider the two functions cos(wt) and sin(wt)
on equal footing. Writing this as
x = real part of A exp(I w t)
pushes the phase into the amplitude A (which is now complex – it has gained a factor exp(I a)) makes
the algebra simpler – especially if you don’t bother to write down the “real part of…” until the end of the
calculation.
Then (for example), we can look for a solution to the forced damped harmonic oscillator
d2 x/dt2 + a dx/dt + b x = F cos(w t)
by putting x = A exp(I wt) (and replacing the right-hand side by F exp(iwt). Then the problem reduces
to algebra (try it).
4. Show that exp(2 pi i) = exp(4 pi i). Show that exp(pi i) = -1, and that exp(I pi/2) = i.
5. Define the logarithm function by ln(exp(x)) = x (that it, it is the inverse of the exponential function),
and assume that it continues to make sense for complex arguments. Then what is ln eix ? What is
ln ei2 pi ? What are some possible values for ln 1, ln(-1), and ln(i)? Give some values for ii .
6. Let a = 2 + 2i, b = 3 + 4i, and c = 2- 2i. Evaluate a*b, a*c, and b/a (hint: multiply b/a by c/c).
7. We can represent a complex number like 3 + 4i as a point on a two-dimensional plane, as show in the
diagram.
Represent the other numbers in Q6 this way.
If we write a = r eix (where r is a real number), what are the values of r and x? What do these represent
on the diagram?
8. The conclusion of Q4 was that ln(z) is a multiply valued function. Show that we can always represent
ln(z) = a + ib
diagram.
where a and b are real, and 0 < b < 2 pi. Explain what this means in terms of the
9. Let’s extend the idea of a derivative to functions of a complex variable by the rule
df/dz = limit as h -> 0
(f(z + h) – f(z))/h
where h is a small complex number. In general this won’t work, because the limit is different when h is
pure real vs when it is pure imaginary. But there are lots of interesting examples where this does work;
functions that have this property are called “analytic functions.”
Show that zn is an analytic function by evaluating the limit as h -> 0 of [(x+iy+ h)n – (x+iy)n)]/h for the
case that h is real and the case that h is imaginary.
Show that the sum of two analytic functions is an analytic function.
Show that exp(z) is an analytic function. What is d/dz exp(z)?
Show that ln(z) is an analytic function, except at z = 0 (where the limit doesn’t make sense). What is
d/dz ln(z)?
10. It is usually true that differentiation is the inverse operation of integration, and usually true that
integration is the inverse of differentiation:
Y
df
dx
dx
X
f (Y )  f ( X )  
The extension of this idea to functions of a complex variable is slightly tricky, but extremely interesting
and useful once we get past the tricky part. Start by claiming the expression above remains true when
X, Y, and x are complex variables and f is a complex-valued function of the complex variable x. You
could use the results of Q9 to find out what the integral of exp(z) and 1/z are. But it would also be
useful to be able to give the usual definition of an integral as being the limit of a sum
Y
 f ( z)dz   f (z )z
i
X
i
i
where the sum is over small pieces of the interval from X to Y. How should we define “the interval from
X to Y” when these are complex? One option is to break up X and Y into real and imaginary parts
(X=a+ib, Y=p+iq), and integrate first over the real part and then over the imaginary part
Y

c
d
a
b
f ( z )dz   f ( x  ib)dx   f (c  iy )idy
X
We get from X to Y along the path shown in the diagram. But another
choice would be to go along the imaginary axis first, and there are
many more ways to get from X to Y. The reason analytic functions
are interesting is that the path doesn’t matter if the function is
analytic everywhere.
Show this by integrating eiz from z=0 to z=1+I along two paths.
11. However, there are lots of functions that are analytic
everywhere except for a few points. Can we ignore the bad spots?
The answer is that the value of
Y
 f ( z )dz
X
is the same for any two paths from X to Y provided that the two paths do not enclose nonanalytic points.
The key example is
Y
1
ln( Y )  ln( X )   dz
z
X
If we start from X=1 (where ln(X) = 0) and go to Y = exp(iy), we will find that the value of the integral is iy.
But then if we go in a circle around the origin, we eventually come to 1 again – but from the point of
view of the logarithm function, this is one of the other ways to write 1, like exp(2 pi i), so that
 1 / z dz  2i
if we go around the origin once in the counterclockwise direction.
We can turn this horrible defect into a wonderful advantage when we observe that if f(z) is analytic
everywhere, f(z)/z is analytic everywhere except for the origin, so we will get the same answer for
 f ( z) / z dz
for all paths that circle the origin once in a counter clockwise direction; consideration of the case of a
very small circle around the origin shows that the result must be
 f ( z) / z dz  2if (0)
The strength of this result is that the answer is the same for any function that is analytic everywhere, for
any path that circles the origin! And by repeating the argument with a small modification, we can show
 f ( z) /( z  a) dz  2if (a)
As an example, consider the integral over all real x of 1/(x2+a2). Taking this to be the real part of
1/(z2+a2) you will see that the function diverges at z = ia and z = -ia; we’ll assume it is analytic elsewhere.
At very large z (including large complex z) the function is very small, so the integral from –infinity to +
infinity along a circle that stays infinitely far from the origin has to be zero (for a semicircle at distance R,
the path length is pi R while the value of the function is of order 1/R2 , so the integral is of order R).
Thus we can turn the integral we are trying to evaluate into an integral over a closed path by integrating
along the real axis and then returning on a semicircle at infinity. According to discussion above, the
value will entirely be due to the enclosed singular point.
Since z2+a2 = (z+ia)(z-ia), the integrand can be represented as the sum of two terms A/(z+ia) and B/(z-ia).
Determine the values for A and B. Then use the rule given above to evaluate the integral, both for the
case that the return path lies in the upper half plane and the case that the return path lies in the lower
half plane.
12. There are many kinds of singularities a function can have. An interesting case is the “simple pole,”
which means that
limit as z  a
(z-a) f(z) = A
is a finite number (so near z=a, f(z) resembles A/(z-a). Then a path integral that goes around the point
z=a (in the counterclockwise direction) will get a contribution 2 pi i A. In this context A is called the
“residue of f(z) at a.”
Find the poles and residues for 1/(z4 + 1)
(the first step is to find solutions to z4 = -1).
13. … well, there’s lots of fun to be had with this subject, but perhaps we should stop here for the time
being.