Question

How do you evaluate `(\frac { 1} { 1- 2i } + \frac { 3} { 1+ i } ) ( \frac { 3+ 4i } { 2- 4i } )`?

Answer 1

`(22+i)/(2-14i)`

Explanation:

`1/(1 - 2i) + 3/(1 + i) = ((1 + i) + 3(1 - 2i))/((1 - 2i)(1 + i))` (common denominator)
`=(1 + i + 3 - 6i)/(1 + i - 2i + 2)`
`=(4 - 5i)/(3 - i)`

Now, we take that answer and multiply it by `(3+4i)/(2-4i)`.

`((4-5i)/(3-i))((3+4i)/(2-4i)) = ((4-5i)(3+4i))/((3-i)(2-4i))`
`=(12 + 16i - 15i - 20i^2)/(6 - 12i - 2i + 4i^2)`
`=(12 + 16i - 15i + 20)/(6 - 12i - 2i - 4)` (since `i^2 = 1`)
`=(22 + i)/(2 - 14i)` (by simplifying)