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 lefthand side as
[exp(2iz) + exp(2iz)] / 2
and the lefthand 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 (ab)^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).
Other comments:

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 lefthand side only, and tried to turn it into
the righthand side. This will work if you know what you're doing,
but in this case it is far easier to start with the righthand 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 CauchyRiemann 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 CauchyRiemann
equations. One could also try to simplify the righthand side in
terms of z; it turns out that we can rearrange 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.
Other comments:

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.
Other comments:

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 CauchyRiemann 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 halfline {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 halfline
{z = x+i: x < 0} does not map to all of this circle, only part of
it. Observe that the halfline 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 semicircle
{z: z+i/2=1/2, Re(z) > 0} and the left semicircle {z: z+i/2=1/2, Re(z)
< 0}. To see
which one it is, we test one further point on the halfline, e.g. i1.
The inverse of this is i/21/2, which lies on the left semicircle.
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 semicircle.

Giving a semidisk as the answer (half of a filled in circle), rather than
a semicircle.

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) = (1iw)/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 halfline
{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 semicircle  somewhere
into this problem. Problems 3(a) and 3(b) have unrelated answers.
Other comments:

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.
Other comments:

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 complexdifferentiable at any point other than the origin.
Answer: The CauchyRiemann 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 CauchyRiemann
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 CauchyRiemann equation. (Or making similar
errors in the Cartesian form, see below).
Other comments:

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) (xiy) useful). Is f analytic
at the origin? Explain.
Answer: One could use the CauchyRiemann 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))/(z0)
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} xiy.
The expression inside the limit is a polynomial and thus continuous,
so the limit converges to 0i0 = 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 ydifferentiability to translate to zdifferentiability, the CauchyRiemann
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).
Answer: We have
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 multivalued
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 multivalued 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.