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

Aug 21, 2017

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

or the $29$ is correct and $k = \frac{87}{10} + \frac{9}{10} i$

#### Explanation:

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

$\left(1 + 3 i\right) \left(2 - 6 i\right) = 2 + 18 = 2 \textcolor{red}{0}$

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:

$\left(1 + 3 i\right) \left(2 - k i\right) = 29$

Divide both sides by $\left(1 + 3 i\right)$ to get:

$2 - k i = \frac{29}{1 + 3 i} = \frac{29 \left(1 - 3 i\right)}{\left(1 - 3 i\right) \left(1 + 3 i\right)} = \frac{29 \left(1 - 3 i\right)}{10}$

Hence:

$k i = 2 - \frac{29 \left(1 - 3 i\right)}{10} = \frac{20 - 29 \left(1 - 3 i\right)}{10} = \frac{- 9 + 87 i}{10}$

So:

$k = \frac{87}{10} + \frac{9}{10} i$