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ENGR 3300
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EXAMINATION III
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Problem 1
Given that the function u = e−y sin x is harmonic, find a function v such that f (z) = u + iv is analytic and express f (z) in
terms of z.
Use the Cauchy-Riemann Equations:
∂u
∂x
= e−y cos x ⇒ v = −e−y cos x + α(x)
∂v
∂y
=
∂v
∂x
= − ∂u
= e−y sin x ⇒ v = −e−y cos x + β(y)
∂y
Therefore, v = −e−y cos x + c where c is a constant that may be carried
along in the following or ignored.
Now, f (z) = u + iv = e−y sin x − ie−y cos x = e−y (sin x − i cos x)
Factor out −i: f (z) = −ie−y (cos x + i sin x) = −ie−y eix = −ie−y+ix
Factor out the i: f (z) = −iei(x+iy) = −ieiz
1
Page 2
Problem 2
6
sin z dz
Evaluate the integral C (z−π/6)
3 if C is the circle |z| = 1 traversed once in the counter-clockwise direction.
H
There is, obviously, a triple pole at z = π/6 inside the circle. Thus we
use n = 2 which means we take the second derivative and divide by 2!=2.
sin6 z
00
= 6 sin5 z cos z
0
= 6 − sin6 z + 5 sin4 z cos2 z
Since π/6 = 30o , we √
know (from a diagram if necessary) that sin(π/6) =
1/2 and cos(π/6) = 3/2 so:
sin
6
00 z √
= 6 −(1/2)6 + 5(1/2)4 ( 3/2)2 = 6(−1+15)/26 = 3·14/32
π/6
Finally, multiplying by (2πi)/2! yields 21πi/16
2
Page 3
Problem 3
Expand
z
(z − 1)(2 − z)
in a Laurent Series centered at z = 2 that converges near that
point z = 2.
f (z) =
First we use partial fractions
z
−z
1
2
=
=
−
(z − 1)(2 − z)
(z − 1)(z − 2)
z−1 z−2
Work on the first term (since the second is done already)
1
1
=
= 1 − (z − 2) + (z − 2)2 − (z − 2)3 · · ·
z−1
(z − 2) + 1
Put them both together to get
f (z) = −
2
+ 1 − (z − 2) + (z − 2)2 − (z − 2)3 · · ·
z−2
which converges in 0 < |z − 2| < 1.
3
Page 4
Problem 4
z−2
Evaluate C z 2(1−2z)
2 dz where the path C is the circle with equation |z − 1| = 43 traversed once in the counter-clockwise direction.
H
There are, obviously, double poles at z = 0 and z = 1/2. The circle C
has center at z = 1 and radius 3/4 so only encloses the z = 1/2 pole.
With n = 1, we must differentiate once and divide by 1! = 1.
First we must factor out the −2 to put the denominator in its correct
form: 4z 2 (z − 1/2)2 . Multiplying the fraction by (z − 1/2)2 we have
z−2
1
−1
−1
1
= 4z
+ 2z
2 . Differentiating yields 4z 2 + z 3
4z 2
Evaluating at z = 1/2 gives −1 + 8 = 7. Finally, multiplying by (2πi)/1!
gives the answer 14πi.
4
Page 5
Problem 5
Evaluate
R∞
0
x4 dx
.
x6 +1
Since the integrand is even, we may calculate, instead,
1 Z ∞ x4 dx
2 −∞ x6 + 1
First we must find the six roots of −1. Since 180o /6 = 30o , the roots in
the upper half plane √
are at 30o , 90o , 150o . The
√ length of these roots must
be one so they are + 3/2 + i/2, i , and − 3/2 + i/2.
The fractional integrand has simple poles so we may just differentiate the
1
z4
denominator: 6z
5 = 6z and substitute values of the poles.
Thus,
"
1
1
1
1
√
√
· (2πi) ·
+
+
2
6( 3/2 + i/2) 6(i) 6(− 3/2 + i/2)
#
Factoring out the 6, we note that the lengths of the two complicated
denominators are both one, so
i
√
1
1h √
πi
π
· (2πi) · ( 3/2 − i/2) − i + (− 3/2 − i/2) = (−2i) =
2
6
6
3
5
Page 6
Problem 6
Find all the values of ii.
By definition, ii = ei ln i . Now ln i = Ln 1 + i( π2 ± 2kπ) = i( π2 ± 2kπ).
π
Then i ln i = −( π2 ± 2kπ) so we have ii = e−( 2 ±2kπ) .
Comment:
π
±k
First note that these are all real numbers! Next, factor it to e− 2 · e2π
Next,
π
a calculator would tell us that e− 2 ≈ 0.20788 while e2π ≈ 535.492. Hence ii is
approximately 0.20788 times any integer multiple of 535.492.
· · · , 6.65 × 10−8 , 0.000388, 0.20788(P.V.), 111.3, 59610, · · ·
Problem 7
What are the residues at each pole of
3iz 4
2
z −iz+2
The quadratic formula gives us the roots of the denominator:
√
i ± −1 − 8
i ± 3i
=
2
2
so there are simple poles at z = 2i and at z = −i. The simplest way to
get residues here is to differentiate the denominator:
3iz 4
2z − i
At z = 2i, this gives
3i(2i)4
3i
= 16. At z = −i, this gives
End of Test
6
3i(−i)4
−3i
= −1.