Chapter_8

# Chapter_8 - VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...

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Unformatted text preview: VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 VIII.1.1 1 2 3 P L K LLL Show that z 4 + 2 z 2 & z + 1 has exactly one root in each quadrant. Solution VIII.1.1 Set p ( z ) = z 4 +2 z 2 & z +1 , and compute 4 arg p ( z ) along the three segments in the sector path in &gure VIII.1.1. First of all p ( z ) have no real zeros. The positive real axis & 1 from to R we parametizize as & x : z = t; ¡ t ¡ R , and our function p ( z ) becomes p ( t ) = t 4 + 2 t 2 & t + 1 = ( t 2 + 1) 2 & t > . Since p ( x ) > for t ¢ , their is no change of argument on this line segment on the real axis, thus 4 arg p ( z ) = 0 . The arc & 2 from R to iR we parametizize as & 2 : z = Re it , ¡ t ¡ ¡= 2 , and our function p ( z ) becomes p ( t ) = R 4 e 4 it + 2 R 2 e 2 it & Re it + 1 . Since the variation in argument is determined by the dominating term R 4 e 4 it for R large the change of argument on this arc is 4 times the variation in argument for the arc, and thus 4 arg p ( z ) £ 4 ¤ & 2 = 2 ¡ . The positive imaginary axis & 3 from iR to we parametizize as & 3 : z = it; ¡ t ¡ R , and our function p ( z ) becomes f ( t ) = ( it ) 4 + 2 ( it ) 2 & it + 1 = ( t 2 & 1) 2 & it . We &nd that the real part have a zero at t = 1 , and the imaginary part have a zero for t = 0 . We make the following table and do the sketches, 2 t Re z Im z arg z 1 + & 1 & & & 2 1 VIII.1.1 (R = 4)-2-1 1 2 3 4-2-1 1 2 3 4 VIII.1.1 (R = 4) Thus when we move from iR to , the argument for p ( z ) remain in the 4-th quadrant except for touching the imaginary axis at & i and terminating at 1 , and their is no change of argument on this line segment on the imaginary axis, thus 4 arg p ( z ) ¡ . Now we have that the total change of argument for p ( z ) is ¡ 2 & , so we have exactly one zero in the &rst quadrat. Because that the roots come in complex conjugate pairs it is plain that p ( z ) have a root in the fourth quadrant too. Becauce that p ( x ) = ( x 2 + 1) 2 & x > it is clear that p ( z ) have no real roots and p ( z ) must have on has one zero in each quadrant. 3 VIII.1.2 1 2 3 P L K Find the number of zeros of the polynomial p ( z ) = z 4 + z 3 +4 z 2 +3 z +2 in each quadrant. Solution VIII.1.2 Set p ( z ) = z 4 + z 3 + 4 z 2 + 3 z + 2 , and compute 4 arg p ( z ) along the three segments in the sector path in &gure VIII.1.2. First of all p ( z ) have no real zeros. The positive real axis & 1 from to R we parametizize as & x : z = t; & t & R , and our function p ( z ) becomes p ( t ) = t 4 + t 3 +4 t 2 +3 t +2 > . Since p ( x ) > for t ¡ , their is no change of argument on this line segment on the real axis, thus 4 arg p ( z ) = 0 . The arc & 2 from R to iR we parametizize as & 2 : z = Re it , & t & ¡= 2 , and our function p ( z ) becomes p ( t ) = R 4 e 4 it + R 3 e 3 it + 4 R 2 e 2 it + 3 Re it + 2 . Since the variation in argument is determined by the dominating term R 4 e 4 it for R large the change of argument on this arc is 4 times the variation in argument...
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Chapter_8 - VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...

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