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

1 Answer
Aug 21, 2017

Either that 2929 should be 2020 and k=6k=6

or the 2929 is correct and k=87/10+9/10ik=8710+910i

Explanation:

If kk is real then in order for the product (1+3i)(2-ki)(1+3i)(2ki) to be real, (2-ki)(2ki) must be a real multiple of (1-3i)(13i), the complex conjugate of (1+3i)(1+3i). Hence k=6k=6, but that gives:

(1+3i)(2-6i) = 2+18 = 2color(red)(0)(1+3i)(26i)=2+18=20

So either there's a typo in the question (understandable since the 00 and 99 keys are just next to one another), or kk is not real.

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

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

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

2-ki = 29/(1+3i) = (29(1-3i))/((1-3i)(1+3i)) = (29(1-3i))/102ki=291+3i=29(13i)(13i)(1+3i)=29(13i)10

Hence:

ki = 2-(29(1-3i))/10 = (20-29(1-3i))/10 = (-9+87i)/10ki=229(13i)10=2029(13i)10=9+87i10

So:

k = 87/10+9/10ik=8710+910i