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

1 Answer
Aug 21, 2017

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#