Simplify (Ci) # (2 - 3i) / (1 + 2i)# and (Cii)# (3i^12-i^9)/(2i+1)#?

1 Answer
Mar 14, 2017

Answer:

# (2 - 3i) / (1 + 2i) = -4/5-7/5i #

# (3i^12-i^9)/(2i+1) = 1/5-7/5i #

Explanation:

We can remove a complex denominator by multiplying the quotient by its complex conjugate (and therefore we must also multiply the numerator), This woks because the product of a complex number with its conjugate is always real, and ia a similar way to rationalising the denominator of an irrational expression to remove a surd denominator.

Question Ci

Here the conjugate of the denominator is #1 - 2i# so we get:

# (2 - 3i) / (1 + 2i) = ((2 - 3i)) / ((1 + 2i)) * ((1 - 2i))/((1 - 2i)) #
# " " = (2-3i-4i+6i^2) / (1+2i-2i-4i^2) #
# " " = (2-7i-6) / (1+4) \ \ \ \ \ (because i^2=-1) #
# " " = (-4-7i) / 5 #
# " " = -4/5-7/5i #

Question Cii
Here the conjugate of the denominator is also #1 - 2i# so we get:

# (3i^12-i^9)/(2i+1) = (3-i)/(1+2i) #
# " " = ((3-i))/((1+2i)) * ((1 - 2i))/((1 - 2i)) #
# " " = (3-i-6i+2i^2)/(1+2i-2i-4i^2) #
# " " = (3-7i-2)/(1+4) #
# " " = (1-7i)/5 #
# " " = 1/5-7/5i #