Question

`(1 + 3i) (2 - ki) = 29` ?

Answer 1

Either that `29` should be `20` and `k=6`

or the `29` is correct and `k=87/10+9/10i`

Explanation:

If `k` is real then in order for the product `(1+3i)(2-ki)` to be real, `(2-ki)` must be a real multiple of `(1-3i)`, the complex conjugate of `(1+3i)`. Hence `k=6`, but that gives:

`(1+3i)(2-6i) = 2+18 = 2color(red)(0)`

So either there's a typo in the question (understandable since the `0` and `9` keys are just next to one another), or `k` is not real.

If `k` is allowed to take complex values then we can proceed as follows:

`(1+3i)(2-ki) = 29`

Divide both sides by `(1+3i)` to get:

`2-ki = 29/(1+3i) = (29(1-3i))/((1-3i)(1+3i)) = (29(1-3i))/10`

Hence:

`ki = 2-(29(1-3i))/10 = (20-29(1-3i))/10 = (-9+87i)/10`

So:

`k = 87/10+9/10i`