Download Complex Analysis: Midterm I Solutions for Math 412 - Prof. Scott Annin and more Exams Mathematics in PDF only on Docsity!
March 2009 Sample Midterm I Name:
Math 412
Problem 1.
(a): Express (1 + i)โ^2 in the form a + bi, where a and b are real numbers.
SOLUTION: We have 1 (1 + i)^2
2 i
i,
so that a = 0 and b = โ
(b): Find Re
4 โ 3 i 2 โ i
SOLUTION: We have
4 โ 3 i 2 โ i
(4 โ 3 i)(2 + i) (2 โ i)(2 + i)
11 โ 2 i 5
i,
so that
Re
4 โ 3 i 2 โ i
(c): Find
4 โ 3 i 2 โ i
SOLUTION: Using the computation in part (b), we have
โฃ โฃ โฃ โฃ
4 โ 3 i 2 โ i
(d): Find
(1 โ i)^2 in polar form.
SOLUTION: Using part (a), we can write
1 (1 โ i)^2
i =
e^3 ฯi/^2.
(Note that the argument of this complex number can be any element of { 3 ฯ/2 + 2ฯk : k โ Z}.)
Problem 2. Find all fourth roots of 16 i.
SOLUTION: We are asked to solve the equation z^4 = 16i for z. Writing
z^4 = 16i = 16ei(^
ฯ 2 +2ฯk) ,
we have z = 2ei(^
ฯ 8 + ฯ 2 k) ,
which we can evaluate for k = 0, 1 , 2 , 3 to find the four roots z 1 , z 2 , z 3 , z 4 :
k = 0 : z 1 = 2ei^ ฯ 8 ,
k = 1 : z 2 = 2ei^ 58 ฯ ,
k = 2 : z 3 = 2ei^ 98 ฯ ,
k = 3 : z 4 = 2ei^ 138 ฯ
.
Problem 3. Prove that if z 6 = 0, then the four points z, iz, โz, and โiz are the vertices of a square with the center at the origin.
SOLUTION: Since |z| = |iz| = | โ z| = | โ iz|, all four points lie on a circle of radius |z| centered at the origin. Moreover, the distance from z to iz is |iz โ z| = |z||i โ 1 | =
2 |z|, the distance from iztoโz is |iz +z| = |z||i+1| =
2 |z|, the distance from โz to โiz is |โz +iz| = |z||โ1+i| =
2 |z|, and the distance from โiz to z is |z + iz| = |z||1 + i| =
2 |z|. Hence, the four points are all equally distant from two neighbors on the circle, thus forming points on a square.
Problem 4. Find the image of the line y = x + 1 under the mapping w = f (z) = iz.
SOLUTION: We have f (z) = f (x + iy) = i(x + iy) = โy + ix. Substituting y = x + 1, we see that the image of this line under f is
f (z) = f (x + iy) = โy + ix = โ(x + 1) + ix = โx โ 1 + ix,
which consists of the points of the form (โx โ 1 , x) in the complex plane. So we have u = โx โ 1 and v = x. That is, u = โv โ 1, so this is the line u + v = โ1.
Problem 5. Find
lim zโ 0
z^2 |z|^2
or show why the limit does not exist.
SOLUTION: Let C 1 denote the path to 0 along the real axis in the complex plane. Thus,
lim zโ 0 on C 1
z^2 |z|^2
= lim xโ 0
x^2 x^2
Next, let C 2 denote the path to 0 along the imaginary axis in the complex plane. Thus,
lim zโ 0 on C 2
z^2 |z|^2
= lim yโ 0
(iy)^2 y^2
Thus, this limit DOES NOT EXIST, since two different paths to 0 resulted in different limiting values.
Problem 6. Using the definition of limit, prove that
lim zโ2+i f (z) = 21 + 19i,
SOLUTION: The domain is the complex plane with origin removed, Cโ. We can create a branch function by restricting the argument of z = eiฮธ^ to some interval ฮฑ < ฮธ โค ฮฑ + 2ฯ, for any ฮฑ โ R. As such, the range of w = z^1 /^3 is the sector of the complex plane consisting of the set
{ฯeiฯ^ : ฯ > 0 and
ฮฑ 3
< ฯ โค
ฮฑ + 2ฯ 3
This is continuous and differentiable except along the branch cut ฮธ = ฮฑ, and intersecting all branch cuts, we see that the only common point is z = 0, so z = 0 is a branch point. This point, however, is excluded from the domain of the function w = z^1 /^3.
