# HW10sol - P ROBLEM SET 10 Partial solution 6 f(z = z 7 6z 3...

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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 ( 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 quarter-circle. 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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