How do you simplify # (5-i)/(2-i) - (3-7i)/(2-3i)# and write in a+bi form?

1 Answer
Dec 3, 2015

Answer:

#8/65 + 64/65 i#

Explanation:

There are two different possibilities:

  • either you combine the two fractions into one first and later write it in the #a + bi# form
  • or you first write each of the fractions in #a + bi# form and later subtract the one from the other

Which approach to pick is a matter of personal preference and doesn't make a big difference.

Let me do the second approach if you don't mind.

1) Transform #(5-i)/(2-i)# into #a + bi# form.

To do so, you need to make use of the formula #(x+y)(x-y) = x^2 - y^2#.
In order to achieve such a situation, you need to multiply both the numerator and the denominator with #2+i#, the conjugate of #2-i#:

#(5-i)/(2-i) = ((5-i)(2+i))/((2-i)(2+i)) = (10 + 3 i - i^2)/(2^2 - i^2)#

... use #i^2 = -1# and simplify...

# = (10 + 3i + 1)/(4 + 1) = (11 + 3i)/5 = 11/5 + 3/5 i#

2) Transform #(3-7i)/(2-3i)# into #a + bi# form.

The approach is the same: multiply the numerator and the denominator with the conjugate of the denominator, namely #2+3i#, and simplify:

#(3-7i)/(2-3i) = ((3-7i)(2+3i))/((2-3i)(2+3i)) = (6 - 5 i - 21 i^2)/(2^2 - (3i)^2) = (6 - 5i + 21)/(4 + 9)#

#color(white)(xxxxx) = (27-5i)/13 = 27/13 - 5/13 i#

3) Subtract the two terms

#(5-i)/(2-i) - (3-7i)/(2-3i) = (11/5 + 3/5 i) -(27/13 - 5/13 i)#

#color(white)(xxxxxxxxxxx) = 11/5 - 27/ 13 + (3/5 + 5/13) i#

#color(white)(xxxxxxxxxxx) = 8/65 + 64/65 i#