MAE294B/SIOC203B: Methods in Applied Mechanics
Winter Quarter 2016
http://web.eng.ucsd.edu/~sgls/MAE294B_2016/
Solution I
1 Show that the geometric locus given by the equation |z − a| + |z + a| =
ellipse by finding its equation in terms of x and y.
Solution. Square both sides, giving
√
a2 + b2 is an
2( x2 + y2 + a2 ) + 2|z − a||z + a| = a2 + b2 .
Now square again, giving
4[( x − a)2 + y2 ][( x + a)2 + y2 ] = [ a2 + b2 − 2( x2 + y2 + a2 )]2 .
This can be simplified to give
x2
y2
+
= 1.
( a2 + b2 )/4 (b2 − 3a2 )/4
This is the equation
for an ellipse
with semi-major and semi-minor
√
√ centered at the origin
1
1
2
2
2
2
2
axis lengths 2 a + b and 2 b − 3a provided b > 3a2 , which can be verified by looking at the original equation on the real and imaginary axes. Note that if b2 < 3a2 , the
equation in x and y corresponds to a hyperbola, whereas there is in fact no solution.
√ This
is an artifact of squaring. The question is better if one uses |z − a| + |z + a| = 2 a2 + b2 ,
which leads to
y2
x2
+
= 1.
a2 + b2 b2
√
This corresponds to an ellipse with semi-major and semi-minor axis lengths a2 + b2 and
b along the x- and y-axes.
2
Find the real and imaginary parts of the following complex functions:
tanh z,
log
z3 − 1
,
z2 + 1
exp (2z2 − ez ).
Solution. First
tanh z =
sinh x cos y + i cosh x sin y
cosh x sinh x + i cos y sin y
sinh ( x + iy)
=
=
.
cosh ( x + iy)
cosh x cos y + i sinh x sin y
sinh2 x + cos2 y
The denominator can also be written cosh2 x − sin2 y. Alternatively
tanh z =
ex+iy − e−( x+iy)
e2x + e2iy − e−2iy − e−2x
sinh 2x + i sin 2y
=
=
.
2iy
−
2iy
2x
−
2x
cosh 2x + cos 2y
e +e +e
+e
ex+iy + e−( x+iy)
1
Next
3
2
2
3
z − 1
z3 − 1
+ i arg z − 1 = 1 log nr + ni + iatan2(ni dr − nr di , nr dr + ni di ),
log 2
= log 2
2
z +1
z + 1
z2 + 1
d2r + d2i
where
nr + ini = x3 − 3xy2 − 1 + i(3x2 y − y3 ),
dr + idi = x2 − y2 + 1 + i2xy
and atan2 is the appropriate way to compute arguments over the whole complex plane.
Finally
exp (2z2 − ez ) = exp [2( x2 − y2 + 2ixy) − ex (cos y + i sin y)]
= e2 ( x
2 − y2 )−ex
cos y
[cos (2xy − ex sin y) + i sin (2xy − ex sin y)].
3 Which of the following functions satisfy the Cauchy–Riemann equations (and where)?
If they satisfy the Cauchy–Riemann equations, give the analytic function of z.
q
ix − y
ix + y
x 2 + y2 ,
,
.
x2 − y2 + 2ixy,
x 2 + y2
x 2 + y2
p
√
x2 + y2 = zz;
Solution. Note that x2 − y2 + 2ixy = z2 . CR satisfied everywhere.
in particular v = 0. CR satisfied nowhere. At the origin, the partial derivatives are not
defined uniquely, e.g. u x = x ( x2 + y2 )−1/2 takes different values as one approaches the
origin along different rays, so can’t use CR there. Near the origin, the full definition of
analyticity leads to the limit of (hh̄)1/2 /h = (h̄/h)1/2 which is not unique for all h → 0,
so the function is analytic nowhere. (ix − y)/( x2 + y2 ) = i/z. CR satisfied nowhere. At
the origin, the partial derivatives are not defined, so the function is analytic nowhere.
(ix + y)/( x2 + y2 ) = i/z. CR satisfied everywhere except at the origin. At the origin, the
partial derivatives are not defined, so the function is analytic everywhere except at the
origin.
H
4 Let C be the unit square with corners at ±1 ± i. Evaluate f (z) dz by computing the
integral explicitly along the contour, where f (z) is given by the following:
1
, (c) z, (d) Re z.
(a) sin z, (b)
2z + 1
Solution. We can do the integral in (a) and (b) explicitly. The integrand in (a) is singlevalued, so the integral vanishes (this is the same as Cauchy’s theorem). In (b), the integral
is equal to the change in 12 log (2z + 1) around the contour. The real part of this returns
to its original value; the imaginary part changes by 2πi, hence the integral is equal to πi
(this is also what the residue theorem would give). For (c) and (d) one has
Z 1
−1
( x + i) dx +
Z 1
−1
(1 − iy)i dy +
Z −1
1
( x − i) dx +
Z −1
1
(−1 − iy)i dy
= 2i + 2i − (−2i) − (−2i) = 8i
and
Z 1
−1
x dx +
Z 1
−1
i dy +
Z −1
1
x dx +
2
Z −1
1
(−1)i dy = 2i − (−2i) = 4i.
5
Show that
Z 2π
dθ
2π
=
.
2
1 − 2t cos θ + t
1 − t2
0
What limits are placed on the parameter t? Why?
Solution. Write z = eiθ . Then the integral can be written as an integral over the unit circle
|z| = 1, where cos θ = 21 (z + z−1 ). Since dz = ieiθ dθ = iz dθ,
I=
Z
|z|=1
dz
1
=
−
1
2
1 − t(z + z ) + t iz
Z
|z|=1
tz2
i
dz.
− (1 + t2 ) z + t
The integrand has poles at the roots of tz2 − (1 + t2 )z + t = (tz − 1)(z − t), i.e. at z = t
and z = t−1 . If |t| < 1, the root z = t is inside the contour, with residue i/(t2 − 1). Hence
T = (2πi) × i/(t2 − 1) = 2π/(1 − t2 ). If |t| > 1, the pole inside the contour is t−1 and
T = 2π/(t2 − 1). (Quick checks: if t = 0, the result is elementary with T = 2π, while
for large t, T ∼ 2πt−2 .) If |t| = 1, the integral does not exist since it has a pole on the
contour (it exists as a Cauchy Principal Value integral). One can avoid using the residue
theorem by decomposing the integrand into partial fractions; each fraction integrates to
a logarithm and tracking the change in the imaginary part of the logarithm as in 4 gives
the result.
6 (Requires Matlab or other numerical/plotting software.) Plot the real and imaginary
parts of the matlab function f(z) = sqrt(1-z.^2). Now repeat for the function f(z) =
z.*sqrt(1./z.^2-1). What is going on? Write a Matlab subroutine to put a branch cut
between −1 and 1 that follows an arbitrary path in the complex plane. Test it for the
path y = 1 − x2 . [Hint: if you do not use Matlab, you will need to investigate how your
software deals with complex logarithms.]
Solution. Figure 1 shows the real and imaginary parts of the two functions f(z) =
sqrt(1-z.^2) and f(z) = z.*sqrt(1./z.^2-1). These may not be the branch cuts you
were expecting. Matlab puts a branch cut where the value of the square root is negative
real, call this − a2 with a ≥ 0. Then in the first case we have 1 − z2 = − a2 , i.e. z2 = 1 + a2 .
This corresponds to branch cuts on the real axis from −∞ to −1 and 1 to ∞. Along the
branch cut, f (z) is pure imaginary so the branch cut shows up only in the imaginary part.
In the second case, z−2 − 1 = − a2 , i.e. z−2 = 1 − a2 . This has real solutions for 0 < a < 1,
giving a branch cut along the real axis from −∞ to −1 and 1 to ∞ that shows up only in
the imaginary part. It also has imaginary solutions for 1 < a, which show up as a branch
cut along the entire imaginary axis that is visible in the real part.
To put in a cut at y = f ( x ) between −1 and 1, we can think of breaking the function
into two square roots, one of z + 1 and of 1 − z. Leave the latter alone. For the former,
decrease its argument by 2π in the regions (x ≤ −1, y ≥ 0), (−1 < x < 1, y > f ( x )) and
(x ≥ 1, y > 0). This choice picks out the branch that is real and positive along the real
axis between −1 and 1. Figure 1 shows plots for this branch cut.
3
Figure 1: (Left) real and (right) imaginary parts of (top) f(z) = sqrt(1-z.^2), (middle)
f(z) = z.*sqrt(1./z.^2-1) and (bottom) the branch with a cut at y = 1 − x2 .
4