Solutions to first midterm, 132/1 Winter 2000

Problem 1 (10 points): Prove that cos(2z) = cos^2(z) - sin^2(z) for all complex numbers z.

Answer: Write the left-hand side as

[exp(2iz) + exp(-2iz)] / 2

and the left-hand side as

([exp(iz) + exp(-iz)] / 2)^2 - ([exp(iz) - exp(-iz)] / 2i)^2.

If you expand both sides and simplify, the two sides will agree, which proves the claim.

Some common errors:

Other comments:
cos(z+w) = cos(z) cos(w) - sin(z) sin(w)
This is true, but it begs the question of why the complex cosine addition law is true.  To be true to the spirit of the question, you would then have to prove the complex cosine addition law, perhaps by breaking up into exponentials.

Problem 2(a) (5 points):  Find an entire function f whose real part is equal to x^2 + xy - y^2.

Answer: Write f = u+iv.  We are given that u(x+iy) = x^2 + xy - y^2.  Since f is entire, we have the Cauchy-Riemann equations

partial u / partial x = partial v / partial y,    partial u / partial y = - partial v / partial x

which in our case simplifies to

partial v / partial y = 2x - y,    partial v / partial x = 2y - x.

Integrating the first equation one gets

v(x+iy) = 2xy - y^2/2 + c(x)

where c(x) is a function that depends only on x (it must be constant in the y direction).  Inserting this into the second equation we get

2y + c'(x) = 2y - x

which solves to

c(x) = -x^2/2 + C

where C is now a genuine constant (something that does not depend on either x or y).  So

v(x+iy) = 2xy - x^2/2 - y^2/2 + C

and so

f(x+iy) = u(x+iy) + i v(x+iy) = x^2 + xy - y^2 + i(2xy - x^2/2 - y^2/2 + C).

Since we only need to find one such function f, we can set C to be any value we want, say 0.  So one possible answer is

f(x+iy) = u(x+iy) + i v(x+iy) = x^2 + xy - y^2 + i(2xy - x^2/2 - y^2/2).

We can then verify that f is analytic by rechecking the Cauchy-Riemann equations.  One could also try to simplify the right-hand side in terms of z; it turns out that we can re-arrange the above as

f(z) = (1 - i/2) z^2

which is clearly entire (it's a polynomial in z).

Some common errors:

Other comments:
partial f / partial x = 1/i  partial f / partial y
to obtain f, but this is trickier as now you have to deal with one complex equation rather than two real equations.  It eventually works, but you have a greater chance of getting confused.

Problem 2(b) (5 points): Explain why there is no entire function f whose real part is equal to y^2 + 2xy.

Answer: The real part of an entire function must be harmonic at every point.  The function u(x+iy) = y^2 + 2xy is not harmonic for any value of x,y:

partial^2 u / partial x^2 + partial^2 u / partial y^2 = 0 + 2  != 0

Thus y^2 + 2xy cannot be the real part of an entire function.

Other comments:

Problem 3(a) (3 points): What is the image of the negative real line {z = x+i0: x < 0} under the map f(z) = 1/(z+i)?

Answer: First, I apologize for using the potentially confusing letter w instead of z to describe the negative real line.  (Strictly speaking, the question is still correctly phrased, just a bit misleading is all).  The i0 term is also just there to emphasize the fact that the z have no imaginary part.  I could just as well have written {z = x: x < 0}.

If you add i to z, the negative real line shifts up one unit, to become the half-line {z = x+i: x < 0}.  This is part of the line {z: Im(z) = 1}.  If you then perform an inversion, the line {z: Im(z) = 1} goes to the circle through the origin with furthest point 1/i = -i, i.e. the circle {z: |z + i/2| = 1/2}.  However, the half-line
{z = x+i: x < 0} does not map to all of this circle, only part of it.  Observe that the half-line goes from i to infinity, so the inverse of this line would go from -i
to 0.  This gives two possibilities for the image: the right semi-circle {z: |z+i/2|=1/2, Re(z) > 0} and the left semi-circle {z: |z+i/2|=1/2, Re(z) < 0}.  To see
which one it is, we test one further point on the half-line, e.g. i-1.  The inverse of this is -i/2-1/2, which lies on the left semi-circle.  So the correct answer is
{z: |z+i/2|=1/2, Re(z) < 0}.  (Question: Why do we exclude the endpoints 0, -i from this circle)?

Common errors:

Problem 3(b) (4 points): What is the inverse image of the negative real line {w = x+i0: x < 0} under the map f(z) = 1/(z+i)?  In other words, what set
maps to the negative real line under f?

Answer: The first step is to invert f.  If w = 1/(z+i), then z = (1/w) - i.  So the inverse transformation is given by

f^{-1}(w) = (1/w) - i

or, if you prefer,

f^{-1}(z) = (1/z) - i.

You can write this transformation also as f^{-1}(w) = (1-iw)/w, but this doesn't make much of a difference.

To find the inverse image of the negative real line under f, we apply f^{-1} to the negative line, i.e. we do an inversion followed by a translation by -i.  If you invert the real line, you get the real line (except for the origin), so if you invert the negative real line, you get the negative real line again (the reciprocal of a negative real number is another negative real number).  If we then shift downwards by -i, we get the half-line {z: Im(z) = -1, Re(z) < 0}, which is the final answer.

Common errors:

Other comments:

Problem 3(c) (3 points): Let Log(z) be the standard branch of the logarithm.  Where is the function Log(1/(z+i)) analytic?

Answer: The function Log(z) is analytic except when z is a negative real number or 0.  So Log(1/(z+i)) is analytic except when z=-i (since then 1/(z+i) is undefined) or when 1/(z+i) is a negative real.  From Problem 3(b), the latter case happens exactly when z lies on the line {z: Im(z) = 1, Re(z) < 0}.  Combining the two together, we see that Log(z) is analytic everywhere except when {z: Im(z) = 1, Re(z) <= 0}.

Common errors:

Other comments:

Problem 4(a) (5 points): Show that the function f(x+iy) = x^2 + y^2 is not complex-differentiable at any point other than the origin.

Answer: The Cauchy-Riemann equation

partial f / partial x = 1/i partial f / partial y
expands out to
2x = (1/i) 2y
which simplifies to
y - ix = 0.
Equating real and imaginary parts, we thus see that the Cauchy-Riemann equation is only satisfied when x=y=0.  So f cannot be differentiable at any point except possibly the origin.

Common errors:

Other comments:
partial u / partial x = partial v / partial y,    partial u / partial y = - partial v / partial x

Problem 4(b) (5 points): Show using first principles that the function f(x+iy) = x^2 + y^2 is complex differentiable at the origin.  (You may find the formula x^2 + y^2 = (x+iy) (x-iy) useful).  Is f analytic at the origin? Explain.

Answer: One could use the Cauchy-Riemann equation(s) to solve this problem, since the partial derivatives of f are indeed continuous, but the question asks for a first principles proof - i.e. using the limit definition of derivative.

By the definition of differentiability, we have to show that the limit

lim_{z -> 0}  (f(z) - f(0))/(z-0)

converges.  Writing z=x+iy and substituting the formula for f, this limit becomes

lim_{x+iy -> 0} (x^2 + y^2) / (x+iy).

Using the given formula, this simplifies to

lim_{x+iy -> 0} x-iy.

The expression inside the limit is a polynomial and thus continuous, so the limit converges to 0-i0 = 0.

f is differentiable at 0, but it is not differentiable on any ball around 0 (thanks to Problem 4(a)).  So f is not analytic at the origin.

Common errors:

Problem 5(a) (5 points)  Let f(z) = exp(1/2  Log(z)) be the principal branch of the square root function z^{1/2}.  Compute f(-i).

Answer: We have

f(-i) = exp(1/2 Log(-i))
Log(-i) = ln |-i| + i Arg(-i).
Since |-i| = 1, ln |-i| = 0.  Also, the principal phase Arg(-i) of -i is equal to -pi/2.  (There are other phases of -i, such as 3 pi / 2 or
3 pi / 2 + 2k pi, but these phases are not in the interval (-pi, pi] and so do not qualify to be the principal phase Arg(-i)).
Log(-i) = - i pi/2
and so
f(-i) = exp(1/2 (-i pi/2) = exp(- i pi/4).
One can also expand this into Cartesian form as
f(-i) = sqrt(2)/2 - i sqrt(2)/2.

Common errors:

Problem 5(b) (5 points) Give an example of a branch g(z) of the square root function z^{1/2} such that g(-i) = exp(3 pi i /4).

Answer: As we see from Problem 5(a), the principal branch exp(1/2 Log(z)) of z^{1/2} does not satisfy all the required properties.  A little experimentation, though, will give other branches which will work.  For instance, the branch

g(z) = exp(1/2 Log_{(0,2 pi]}(z) )

will work.  (I apologize for the inability to typeset mathematical subscripts etc. in HTML, but I hope the above is readable).  This is indeed a branch of the square root function, and if one evaluates it at -1 one obtains exp(3 pi i/4) as desired, because

Log_{(0,2pi]} (-1) = ln |-1| + i Arg_{(0,2pi]}(-1) = 0 + i (3 pi /2)

- note that 3 pi /2 is the only phase of -1 which lies in the range (0,2pi].

Common errors: