1. (a)
1 + j=√2ej(arctan 1
1)=√2ej(1
4π).
(b)
(1 + j)25 =√2ej(1
4π)25 =√225
ej(25
4π)=√225
ej(6π+1
4π)=√225
ej(1
4π).
(c)
ez+1 = 1 + j=√2ej(1
4π)=eln √2+j(1
4π)=eln 2
2+j(1
4π)
per tant
Re(z+ 1) = Reln 2
2+j(1
4π)(1)
Im(z+ 1) = Imln 2
2+j(1
4π)+2πkj ;k∈Z(2)
De (1) tenim
Re(z) + 1 = ln 2
2→Re(z) = ln 2
2−1
de (2) temin
Im(z) = π
4+ 2πk
per tant
z=ln 2
2−1 + jπ
4+ 2πk;k∈Z
2. p(z) = z4+ 4
(a) p(z)=0.
z4=−4→z=4
√−4 = 4
√4ejπ =√2ej(π
4+π
2k), k = 0,1,2,3.
(b)
p(z) = z−√2ej(π
4)z−√2ej(3π
4)z−√2ej(5π
4)z−√2ej(7π
4)
= (z−(1 + j))(z−(−1 + j))(z−(−1−j))(z−(1 −j)).
(c)
p(z) = z2+ 2z+ 2z2−2z+ 2.
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