February 23-27, 2009 Week #5 Practice Quiz Name:
Math 412
SOLUTIONS
Problem 1. (5 points) Evaluate
lim
z→−i
z6+ 1
z2+ 1.
SOLUTION: Direct substitution of z=−iis impossible here, since both numerator and denom-
inator evaluate to zero. This suggests the use of L’Hopital’s Rule. Setting f(z) = z6+ 1 and
g(z) = z2+ 1, we find that f0(z) = 6z5and g0(z) = 2z. Thus,
lim
z→−i
z6+ 1
z2+ 1 = lim
z→−i
6z5
2z= lim
z→−i3z4= 3(−i)4= 3.
2
Problem 2. Let
f(z) = z2+ 1
(z+ 2)(z2+ 2z+ 2).
(a): (3 points) Determine all singular points of f.
SOLUTION: Singular points occur where the derivative does not exist. For a rational function,
this is precisely where the denominator is zero. The denominator is zero precisely when z=−2
or when z2+ 2z+ 2 = 0. Via the quadratic formula, we find that z2+ 2z+ 2 = 0 if and only if
z=−1±i. Hence, the singular points of fcomprise the set {−2,−1 + i, −1−i}.2
(b): (3 points)Determine f0(z0)for a non-singular point z0∈C.
SOLUTION: Using the quotient and product rules for differentiation, we have
f0(z0) = (z0+ 2)(z2
0+ 2z0+ 2)(2z0)−(z2
0+ 1) (z0+ 2)(2z0+2)+(z2
0+ 2z0+ 2)
(z2
0+ 2z0+ 2)2.
I beg you: don’t simplify this! 2
Problem 3. (6 points) Determine, with proof, a formula for f0(z0), where f(z) = z5−z2.
SOLUTION: We have
f0(z0) = lim
z→z0
(z5−z2)−(z5
0−z2
0)
z−z0
= lim
z→z0
z5−z5
0
z−z0
−lim
z→z0
z2−z2
0
z−z0
= lim
z→z0
(z−z0)(z4+z3z0+z2z2
0+zz3
0+z4
0)
z−z0
−lim
z→z0
(z−z0)(z+z0)
z−z0
= lim
z→z0
(z4+z3z0+z2z2
0+zz3
0+z4
0)−lim
z→z0
(z+z0)
= 5z4
0−2z0.
