BRIEF QUIZ ANSWERS
In this document you will find brief answers to quiz questions.
If the quiz question is a homework problem, and the answer
is in the book, then the brief answer that you
will see below will be only ``Answer is in the book.''
QUIZ
Week 1
Week 2
- (1/5) + (13/5)i
-
- a circle centered at 1+i with radius square_root(2)
- a circle centered at 0 with radius 1
- Choose w so that w^2 = z. We must show that |w| = square_root(|w^2|).
This holds because square_root(|w^2|) = square_root(|w|^2) = |w|.
(Here we have used that the square root of the square of
a positive real number |w| is equal to |w|.)
- z = conjugate(z) holds in R but fails in C (since it
fails for z = i).
Week 3
- cos(Pi/6) + i*sin(Pi/6), cos(5*Pi/6) + i*sin(5*Pi/6),
cos(3*Pi/2) + i*sin(3*Pi/2).
-
- Let A and B be open, and let C be the intersection
of A and B. If p is an arbitrary point of C, then p is in both
A and B. These sets are open, so p is in the
interior of A and also of B.
This means that there is an neighborhood of radius
R of p that lies in A, and a neighborhood of radius
R' of p that lies in B. The neighborhood with the smallest
radius lies in C, so p is an interior point of C.
Since p was an arbitrarily chosen point in C, all points
of C are interior points, so C is open.
- u = (x^2+x+y^2)/((x+1)^2+y^2), v = y/((x+1)^2+y^2)
Week 4
- Derivative = 2y - 2ix.
- Must show that f^2(z) is differentiable everywhere.
By the chain rule, [f^2(z)]' = 2*f(z)*f'(z), which exists
everywhere since f'(z) does.
- f(z) = |z|^2 = x^2 + y^2, so u = x^2 + y^2 and v = 0.
u_x = v_y iff 2x = 0, and u_y = -v_x iff 2y = 0.
Both hold only when x = y = 0. Thus, the only z_0
is z_0 = 0.
- u_x = v_y = 2x-2y. u_y = -v_x = -2x-2y.
Week 5
-
- Pi*i + 2*n*Pi*i
- Pi*i + 2*n*Pi*i
- By Duncan Macleod's Corollary,
The only entire function that agrees with f(x) = |x^3|
on the positive real axis is F(z) = z^3. Similarly, the
only entire function that agrees with f(x) = |x^3|
on the negative real axis is G(z) = -z^3. Since F(z) is
not equal to G(z), there is no entire function
that agrees with f(x) on the whole real line.
Week 6
Week 7
No quiz.
Week 8
- (Pi/2)*i
- 2*a*b*i
- sinh(2*Pi)
- Any circle that does not contain z = 0 lies in
some half plane D. On D, any branch of
z^(1/2) has an antiderivative (equal to some branch of
(2/3)*z^(3/2)). It follows that the integral if z^(1/2)
around any loop in D is zero.
Week 9
- 0
- 2*Pi*i
- -(8/9)*Pi*i
- If f(z)-g(z) is zero on C, then for any point z_0 inside
C we have f(z_0) - g(z_0) = the integral over C of [f(z) - g(z)]/(z-z_0)
= 0. Thus f(z_0) = g(z_0) for any z_0 inside C.
Week 10
- Use the Cauchy Integral Formula to estimate |f''(z_0)|.
The estimate that arises from integrating around
the circle |z|=r with r greater than |z_0| is
Mr^2/(r-|z_0|)^3, which goes to zero as r gets large.
Thus f''(z) = 0 for all z. You get f(z)=az+b by
integrating.
- If u is harmonic on and inside C, then it
has a harmonic conjugate v on and inside C.
The function exp(u+iv) is analytic on and inside
C, so its modulus |exp(u+iv)| = exp(u) assumes
a maximum on C. Since exp(x) is an increasing function
of a real variable, and u is real valued, this implies
that u has a maximum on C.
Week 11
Test 2.
Week 12
No quiz.
Week 13
- f(z) = sin(z)/z + 1/(z-1) + exp(1/(z-2)) if z is not 0,
and f(0) = 0. This function has a removable singularity at z=0,
a pole at z = 1, and an essential singularity at z = 2.
- ... + 1/(3z^4) - 1/(3z^3) + 1/(3z^2) - 1/(3z) - 1/6 - z/12 - z^2/24 - z^3/48 - z^4/96 - ...
- 1/3.
Week 14
Week 15
Last modified on Sep 6, 2000.