Hoe bereken ik de reële en imaginaire delen van deze vergelijking?

Antwoord:

# "Echt deel" = 0,08 * e ^ 4 #

# "en Imaginary part" = 0,06 * e ^ 4 #

Uitleg:

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

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

# => e ^ (2 + i * pi / 2) = e ^ 2 * exp (i * 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 - 0.3 i #

#"Dus we hebben"#

# (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) #

# => "Echt deel" = 0,08 * e ^ 4 #

# "en Imaginary part" = 0,06 * e ^ 4 #

Antwoord:

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

Uitleg:

Herhaal dat, # E ^ (itheta) = costheta + isintheta ………….. (vierkant) #.

#:. 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 en Im (z) = 3 / 50e ^ 4 #.

Antwoord:

# #

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

Uitleg:

# #

# "We zullen dit oplossen, werken aan de complexe exponentiële" #

# "deel eerst." #

# "Daar gaan we: " #

# ({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 = (e ^ {4} (- 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 = e ^ 4 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 kleur (rood) annuleren {2} cdot (4 +3 i) / {kleur (rood) annuleren {2} cdot 50} = e ^ 4 cdot (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. #

# #

# "Dus:" #

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