February 27, 2009 Quiz #2 Name:
Math 412
SOLUTIONS
Problem 1. Let f(z) = (12 + 5i)z+ 2 −3i.
(a): (5 points) Find
lim
z→2+if(z).
SOLUTION: Since fis a polynomial, it is continuous on all of C. Therefore, limits
can be evaluated by “plugging in” the point z0= 2 + i:
f(2 + i) = (12 + 5i)(2 + i)+2−3i= 21 + 19i.
2
(b): (5 points) Let g(z) = f z2−2zi
z2+ 4 !. Find all points z∈Csuch that gis
continuous. Write a very brief justification. (Don’t give an (, δ)proof.)
SOLUTION: The function h(z) = z2−2zi
z2+ 4 is a rational function, hence continuous
every that the denominator is nonzero. Thus, his continuous on C− {2i, −2i}. The
function fis continuous throughout C. Since the composition of continuous functions
remains continuous, we conclude that gis continuous on precisely the same set on
which his continuous:
C− {2i, −2i}.
2
Problem 2. (4 points) Let α=π/2. Define fα(z) = z1/4=4
q|z|eiθ/4, where
π/2< θ ≤5π/2. Find the range of the function w=fα(z).
SOLUTION: Since zhas arbitrary modulus, 4
q|z|can take any value in R+. Since
π/2< θ ≤5π/2,
we have
π/8< θ/4≤5π/8.
Hence, the range of fαconsists of the sector of the complex plane with points whose
polar form has arbitrary modulus and angle lying in the interval π
8,5π
8i.2
