MAE294B/SIO203B: Methods in Applied Mechanics
Winter Quarter 2010
http://maecourses.ucsd.edu/mae294b
Solution VIII
1
Find the points where the radicals vanish:
z
2
+
1
=
0 gives
±
i; 2

√
z
2
+
1
=
0 gives
±
√
3.
There are hence 4 branch points and many ways of joining them. The default Matlab branch cut
takes
z
2
+
1
=

u
2
with
u
>
0, i.e. the imaginary axis with

y

>
1 and
z
2
+
1
= (
2
+
v
2
)
2
with
v
>
0, i.e. the real axis with

x

>
√
3. There are potentially subtle issues if a branch of
√
z
2
+
1 is
chosen that is everywhere negative; ignore these. Figure 1 shows the real and imaginary parts of
two different branches computed by Matlab.
Figure 1: (a) and (b): real and imaginary parts of
sqrt(2  sqrt(z.^2+1))
. (c) and (d): real and
imaginary parts of
sqrt(2  i sqrt(z.^21))
.
2
We have to choose a branch of
(
z
2

1
)
1
/
2
.
The obvious one has a cut along the real axis
between

1 and 1, and is positive and real on the real axis with
x
>
1. Consider the integral
Z
C
d
z
√
z
2

1
around a contour
C
surrouding the cut. Shrinking the contour onto the cut gives
Z
1
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 Spring '09
 MAE294B
 Complex number, imaginary parts, real axis, Applied Mechanics

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