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

1 Answer
Aug 21, 2017

Either that 29 should be 20 and k=6

or the 29 is correct and k=8710+910i

Explanation:

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

(1+3i)(26i)=2+18=20

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)(2ki)=29

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

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

Hence:

ki=229(13i)10=2029(13i)10=9+87i10

So:

k=8710+910i