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 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
Solutions to first midterm, 132/1 Winter 2000
1 of 7
9/19/2011 3:46 PM
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View Full Documentv(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:
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 Spring '08
 Grossman
 Math, Complex Numbers, Complex number

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