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P
ROBLEM SET
10
Partial solution
6.
f
(
z
) =
z
7
+6
z
3
+7
. Consider the curve
γ
from Example 2, for
R
large. On the segment
from
0
to
R
on the real axis
f
(
x
) =
x
7
+ 6
x
3
+ 7
,
f
0
(
x
) = 7
x
6
+ 18
x
2
≥
0
. Since
f
(0) = 7
>
0
the function
f
is positive on
0
≤
x
≤
R
and hence the argument is equal
to zero.
On the quarter circle
z
=
Re
it
,
0
≤
t
≤
π/
2
, for
R
large the argument of
f
(
Re
it
)
is
approximately equal to the argument of
R
7
e
7
it
. Hence the argument of
f
(
z
)
changes by
about
7
π/
2
over the quartercircle.
On the line segment from
iR
to
0
along the imaginary axis
z
=
iy
where
0
< y
≤
R
;
f
(
iy
) =

iy
7

6
iy
3
+7
. Note that real part of
f
(
iy
)
is positive, while the imaginary part
in negative. Hence for any
y >
0
f
(
iy
)
is in the fourth quadrant. Hence the argument
cannot change by more then
π/
2
, in particular
f
(
iy
)
can not make any loops around the
origin as
y
goes from
R
to
0
. Since the total change over the curve
γ
has to be a multiple
of
2
π
this mens that the total change of argument over
γ
is
4
π
.
Hence by the argument principle (since
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This document was uploaded on 03/27/2010.
 Spring '09

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