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

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

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Answer 1 · 2017-08-21T23:18:13

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

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

Explanation:

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

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

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

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

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

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

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

Hence:

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

So:

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

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

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

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