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 from0 to R on the real axis f ( x ) = x 7 + 6 x 3 + 7 , f0 ( x ) = 7 x 6 + 18 x 2 ≥0 . Since f (0) = 7 >0 the function f is positive on0 ≤ 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 quarter-circle. On the line segment from iR to0 along the imaginary axis z = iy where0 < 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 to0 . 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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