## 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:

• Expanding (a+b)^2 as a^2 + b^2, or (a-b)^2 as a^2 - b^2.
• Forgetting about various i, especially in the denominator in the definition of sin(z).
• Interchanging the definitions of sin(z) and cos(z).
• Some of you showed that the identity followed from the complex cosine addition law
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.
• Some of you tried to expand cos(z) = cos(x+iy) using the cosine addition law.  This is not exactly an error, but it is not really progress either.  It also begs the question as before.
• Some of you expanded the left-hand side only, and tried to turn it into the right-hand side.  This will work if you know what you're doing, but in this case it is far easier to start with the right-hand side and expand that.  (Actually, for these problems it is usually best to expand out both sides and try to get them to meet somewhere in the middle).

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:

• Some people put things like 2xy - x^2/2 - y^2/2 + C as the answer.  This isn't f, this is v.  Read the question carefully!
• Mixing up the u's, v's, x's, y's and plus/minus signs was somewhat common.
• One can also use the alternate form
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.

• It is possible for u to be harmonic for some values of x,y and not harmonic for other values of x,y.  Then u still cannot be the real part of an entire function, though it could possibly be a real part of an analytic function defined on a domain where u is harmonic.
• One could also solve the Cauchy-Riemann equations and show that they lead to no solution.

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:

• Giving the full circle as the answer, rather than a semi-circle.
• Giving a semi-disk as the answer (half of a filled in circle), rather than a semi-circle.
• Assuming the answer was a straight line.  (If you take a line through the origin and apply an inversion w=1/z, you get another line through the origin.  But this map isn't the inversion map, it's w = 1/(z+i)).
• Plotting 1/i above the real axis (1/i = -i lies below the unit axis), which generally led to confusion, especially if one was trying to guess the answer by plotting two or three points.
• Plotting the graph of the real function y=1/x or y=1/x+1.  Such graphs have absolutely nothing to do with this problem.
• Inverting the map 1/(z+i).  (See Problem 3(b) below).

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:

• Failing to invert the map 1/(z+i).  (See Problem 3(a) above).
• Computing the inverse of 1/(z+i) as z+i.  (Despite the fact that the map w=1/z is known as the inversion map, the concept of an inverse of a function is distinct from the notion of the reciprocal of a function.)
• Inserting the answer from Problem 3(a) - i.e. the semi-circle - somewhere into this problem.  Problems 3(a) and 3(b) have unrelated answers.
• Some of you obtained the answer in parametric form, e.g. { z = 1/x - i: x a negative real }.  For this particular case, this type of answer is reasonable.  In other questions you may be asked to sketch the set, in which case a parametric answer is insufficient.

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:

• Believing that Log(z) is analytic everywhere except at 0.  This isn't true, because of the branch cut caused by the Arg function.
• Stating that Log(1/(z+i)) is analytic everywhere except at 0 and the negative real axis.  That's where Log(z) fails to be analytic, but Log(z) is not the same function as Log(1/(z+i)).
• Writing Re(z) < 0 as z < 0.  Or, saying that Log(z) is only analytic "when z > 0".  Neither of these statements even make sense.  One cannot compare one complex number with another complex number.  If you ever get to the point when you're writing things like z > w, you've done something wrong.
• Interestingly, several people were able to answer 3(c) correctly without doing 3(b) correctly!  The intent was to use 3(b) to solve 3(c).

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.

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:

• Forgetting the 1/i in the Cauchy-Riemann equation.  (Or making similar errors in the Cartesian form, see below).
• One can also use the Cartesian form
partial u / partial x = partial v / partial y,    partial u / partial y = - partial v / partial x
to achieve the same conclusion.  Note that in this case u(x+iy) = x^2 + y^2 and v(x+iy) = 0.
• Another approach is to show that the real part of f (which is again x^2 + y^2) is not harmonic at any point.  This shows that f is not analytic anywhere, but does not preclude the possibility that it is differentiable at a few isolated points.

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:

• Observing that f was differentiable with respect to x and y, and then concluding that f must be differentiable with respect to z.  (In order for x- and y-differentiability to translate to z-differentiability, the Cauchy-Riemann equations must be satisfied).
• Assuming that, since f is differentiable at the origin, that f must also be analytic at the origin.  Analyticity is a stricter property than differentiability.

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).

f(-i) = exp(1/2 Log(-i))
and
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)).
So
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:

• Computing the principal phase of -i as 3pi/2, 3pi/2 + 2k pi, or -pi/2 + 2k pi.  These are all phases of -i, but they are not the principal phase.
• Writing f(-i) as (-i)^{1/2}.  Strictly speaking, (-i)^{1/2} is a multi-valued quantity (it can take the values exp(- i pi/4) and exp(3 pi i/4)); f(-i) is the principal branch p.v.(-i)^{1/2} of this multi-valued quantity.  Many of you then manipulated (-i)^{1/2}, e.g. as i^{3/2}; these answers are still incorrect.  f(-i) is a single number, not a range of multiple values, as the quantities (-i)^{1/2} or i^{3/2} suggest.
• Deciding that the answer exp(3pi i/4) was correct because 3pi/4 lies in the interval (-pi, pi].  This is not how principal branches work.  Certainly the principal logarithm Log(-i) of -i is required to have its imaginary part lie in the interval (-pi, pi]; however, the quantity 3pi/4 arises as the (potential) imaginary part of half of Log(-i), not Log(-i) itself.

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:

• Many of you differentiated g(z) rather than specified what g(z) should be.  This was not what the question asked!
• Some of you also worked out where g(z) was analytic.  Again, this is not what the question asked.  Read the question carefully.
• In many cases g(z) was not written out correctly.  Answers like (0,2 pi] or Log_{(0,2 pi]}(z) or 1/2 Log{(0,2 pi]}(z) are not correct.  The correct description of the branch g(z) is as given above.
• In some cases functions were described which were not branches of the square root function at all!  For instance, the function exp(2 Log(z)) is not a branch of z^{1/2} (it does not evaluate to a square root of z, in fact it evaluates to z^2) and is an unacceptable answer.