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380_hw_2_solutions[1]

# 380_hw_2_solutions[1] - 1.4.2b 2e3 i =6 Math 380 Solutions...

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Math 380 Solutions to selected problems from hw 2 1.4.2b 2 e 3+ i°= 6 = 2 e 3 (cos( °= 6) + i sin( °= 6)) = 2 e 3 ( p 3 + i ) = 2 = e 3 p 3 + ie 3 : 1.4.4b 2 + 2 i ° p 3 + i = 2 p 2 e i°= 4 2 e ° i 5 °= 6 = p 2 e 13 °i= 12 = ° p 2 e i°= 12 : 1.4.10 Suppose z = x + iy with x ± 0 : Then j e z j = ° ° e x e iy ° ° = j e x j ° ° e iy ° ° = e x : Since x ± 0 , it follows that e x ± 1 ; so j e z j ± 1 : 1.5.10 The roots of the equation z 4 +1 = 0 are e i°= 4 ; e 3 °i= 4 ; e 5 °i= 4 ; and e 7 °i= 4 : The °rst and last roots form a conjugate pair as do the two middle roots. Thus, we can reorder the roots and rewrite them as e i°= 4 ; e ° °i= 4 ; e 3 °i= 4 ; and e ° 3 °i= 4 : Thus z 4 + 1 = [( z ° e i°= 4 ) ± z ° e ° i°= 4 ² ][( z ° e i 3 °= 4 ) ± z ° e ° i 3 °= 4 ² ] = h z 2 ° z ( e i°= 4 + e ° i°= 4 ) + 1 i h z 2 ° z ( e i 3 °= 4 + e ° i 3 °= 4 ) + 1 i = h z 2 ° 2 z Re e i°= 4 + 1 i h z 2 ° 2 z Re e i 3 °= 4 + 1 i = " z 2 ° 2 z p 2 2 + 1 # " z 2 ° 2 z ° p 2 2 ! + 1 # = h z 2 ° p 2 z + 1 i h z 2 + p 2 z + 1 i : 1.6.2 I can±t sketch with this program, so I±ll describe each region.

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