Bagaimanakah saya mengira Bahagian Sebenar dan Imaginary persamaan ini?

Jawapan:

# "Bahagian sebenar" = 0.08 * e ^ 4 #

# "dan sebahagian Imaginer" = 0.06 * e ^ 4 #

Penjelasan:

#exp (a + b) = e ^ (a + b) = e ^ a * e ^ b = exp (a) * exp (b)

#exp (i theta) = cos (theta) + i sin (theta) #

(= pi / 2) = e ^ 2 * (cos (pi / 2) + i sin (pi / 2)) #

# = e ^ 2 * (0 + i) = e ^ 2 * i #

# 1 / (1 + 3i) = (1-3i) / ((1-3i) (1 + 3i)) = (1-3i) / 10 = 0.1 -

# "Jadi kami ada" #

# (e ^ 2 * i * (0.1-0.3 i)) ^ 2 #

# = e ^ 4 * (- 1) * (0.1-0.3 * i) ^ 2 #

# = - e ^ 4 * (0.01 + 0.09 * i ^ 2 - 2 * 0.1 * 0.3 * i) #

# = - e ^ 4 * (-0.08 - 0.06 * i) #

# = e ^ 4 (0.08 + 0.06 * i) #

# => "Sebenarnya" = 0.08 * e ^ 4 #

# "dan sebahagian Imaginer" = 0.06 * e ^ 4 #

Jawapan:

# Rl (z) = 2 / 25e ^ 4, dan, Im (z) = 3 / 50e ^ 4 #.

Penjelasan:

Ingatlah bahawa, # e ^ (itheta) = costheta + isintheta ………….. (persegi) #.

#:. z = ((e ^ (2 + ipi / 2)) / (1 + 3i)) ^ 2 #, # = (e ^ (2 + ipi / 2)) ^ 2 / (1 + 3i) ^ 2 #, # = e ^ (2 * (2 + ipi / 2)) / (1 + 3i) ^ 2 #, # = e ^ (4 + ipi) / (1 + 3i) ^ 2 #, # = (e ^ 4 * e ^ (ipi)) / (1 + 3i) ^ 2 #, # = {e ^ 4 * (cospi + isinpi)} / (1 + 3i) ^ 2 #,

# = {e ^ 4 (-1 + i * 0)} / (1 + 3i) ^ 2 #, # = - e ^ 4 * 1 / (1 + 3i) ^ 2 * (1-3i) ^ 2 / (1-3i) ^ 2 #, # = - {e ^ 4 (1-3i) ^ 2} / {(1 + 3i) (1-3i)} ^ 2 #, # = - {e ^ 4 (1-3i) ^ 2} / (1-9i ^ 2) ^ 2 #, # = - (e ^ 4 (1-6i + 9i ^ 2)) / {1-9 (-1)} ^ 2 #, # = - (e ^ 4 (1-6i-9)) / (10) ^ 2 #, # = - (e ^ 4 (-8-6i)) / 100 #, # = (e ^ 4 (4 + 3i)) / 50 #.

#rArr Rl (z) = 2 / 25e ^ 4, dan Im (z) = 3 / 50e ^ 4 #.

Jawapan:

# #

# qquad qquad qquad qquad qquad quad ({e ^ {2 + i pi / 2}} / {1 + 3 i}) ^ 2 = {2 e ^ 4} / 25 + {3 e ^ 4} / 50 i. #

Penjelasan:

# #

# "Kami akan melakukan ini, bekerja pada eksponen kompleks" #

# "bahagian pertama." #

# "Di sini kita pergi:" #

# ({e ^ {2 + i pi / 2}} / {1 + 3 i}) ^ 2 = (e ^ {2 + i pi / 2}) ^ 2 / (1 + 3 i) ^ 2 = (e ^ {4 + i pi}) / (1 + 3 i) ^ 2 = (e ^ {4} e ^ {i pi}) / (1 + 3 i) 2 #

# qquad qquad qquad = {e ^ {4} (cos (pi) + i sin (pi))} / (1 + 3 i) ^ 2 1 + i cdot 0)) / (1 + 3 i) ^ 2 #

# qquad qquad qquad = e ^ 4 cdot {-1} / (1 + 3 i) ^ 2 = e ^ 4 cdot {-1} / (1 + 3 i) ^ 2 cdot (1 - 3 i) ^ 2 / (1 -3 i) ^ 2 #

# qquad qquad qquad = e ^ 4 cdot {-1 cdot (1 - 3 i) ^ 2} / {(1 + 3 i) ^ 2 (1 -3 i) ^ 2} e ^ 4 cdot {-1 cdot (1 - 3 i) ^ 2} / {(1 + 3 i) (1 -3 i) ^ 2}

# qquad qquad qquad = e ^ 4 cdot {-1 cdot (1 - 6 i + 9 i ^ 2)} / (1 ^ 2 + 3 ^ 2) ^ 2 cdot {-1 cdot (1 - 6 i - 9)} / 10 ^ 2 #

# qquad qquad qquad = e ^ 4 cdot {-1 cdot (-8 - 6 i)} / 100 = e ^ 4 cdot {8 + 6 i} / 100 #

# qquad qquad qquad = e ^ 4 cdot warna (merah) batalkan {2} cdot (4 +3 i) / {color (red) 4/50 +3/50 i) #

# qquad qquad qquad = e ^ 4 cdot (2/25 +3/50 i) = {2 e ^ 4} / 25 + {3 e ^ 4} / 50 i. #

# #

# "Jadi:" #

# qquad qquad qquad qquad qquad qquad ({e ^ {2 + i pi / 2}} / {1 + 3 i}) ^ 2 = {2 e ^ 4} / 25 + {3 e ^ 4} / 50 i. #