Što je jednako (1-3i) / sqrt (1 + 3i)?

Odgovor:

# (1-3i) / sqrt (1 + 3i) #

# = (- 2sqrt ((sqrt (10) 1) / 2) + 3 / 2sqrt ((sqrt (10) 1) / 2)) - (2sqrt ((sqrt (10) 1) / 2) + 3 / 2sqrt ((sqrt (10) 1) / 2)) i #

Obrazloženje:

Općenito, kvadratni korijeni # A + bi # su:

# + - ((sqrt ((sqrt (a ^ 2 + b ^ 2) + a) / 2)) + (b / abs (b) sqrt ((sqrt (a ^ 2 + b ^ 2) -a) / 2)) i) #

Vidi:

U slučaju # 1 + # 3i, i Real i imaginarni dijelovi su pozitivni, tako da je u Q1 i ima dobro definiran glavni korijen:

#sqrt (1 + 3i) #

# = sqrt ((sqrt (1 ^ 3 ^ 2 + 2) + 1) / 2) + sqrt ((sqrt (1 ^ 3 ^ 2 + 2) -1) / 2), i #

# = sqrt ((sqrt (10) +1) / 2) + sqrt ((sqrt (10) -1) / 2) i #

Tako:

# (1-3i) / sqrt (1 + 3i) #

# = ((1-3i) sqrt (1 + 3i)) / (1 + 3i) #

# = ((1-3i) ^ 2 sqrt (1 + 3i)) / ((1 + 3i) (1-3i)) #

# = ((1-3i) ^ 2 sqrt (1 + 3i)) / 4 #

# = 1/4 (1-3i) ^ 2 (sqrt ((sqrt (10) +1) / 2) + sqrt ((sqrt (10) -1) / 2) i) #

# = 1/4 (-8-6i) (sqrt ((sqrt (10) +1) / 2) + sqrt ((sqrt (10) -1) / 2) i) #

# = - 1/2 (4 + 3i) (sqrt ((sqrt (10) +1) / 2) + sqrt ((sqrt (10) -1) / 2) i) #

# = - 1/2 ((4sqrt ((sqrt (10) 1) / 2) -3sqrt ((sqrt (10) 1) / 2)) + (4sqrt ((sqrt (10) 1) / 2) + 3sqrt ((sqrt (10) 1) / 2)) i) #

# = (- 2sqrt ((sqrt (10) 1) / 2) + 3 / 2sqrt ((sqrt (10) 1) / 2)) - (2sqrt ((sqrt (10) 1) / 2) + 3 / 2sqrt ((sqrt (10) 1) / 2)) i #