Answer:
(2 +4i)(5 -6i) . . . or . . . (4 -2i)(6 +5i)
Explanation:
The product of two complex numbers is ...
(a +bi)(c +di) = (ac -bd) +(bc +ad)i
So, we're looking for pairs of numbers that can be combined in different ways to give 34 and 8. The numbers we found (by trial and error) are ...
2, 4, 5, 6
where 4*6 +2*5 = 34 and 4*5 -2*6 = 8. Because of the effect if i^2 on the sign, we need to have the imaginary parts have opposite signs.
Each of the solutions shown above is representative of 4 solutions. For example, for the first one, you could have ...
(2 +4i)(5 -6i) = (2·5 +4·6) + (4·5 +2(-6))i = 34 +8i
(5 -6i)(2 +4i) = (5·2 +6·4) + (-6·2 +5·4) = 34 +8i . . . . . order of factors swapped
(-2 -4i)(-5 +6i) = ((-2)(-5) -(-4)(6)) + ((-4)(-5) +(-2)(6))i = 34 +8i . . . . both factors in the first solution negated
(-5 +6i)(-2 -4i) = ((-5)(-2) -(6)(-4)) +(6(-2) +(-5)(-4))i = 34 +8i . . . . factors swapped and negated
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Likewise, the second shown solution above is representative of 4 solutions.
Possible solutions are ...
- (2 +4i)(5 -6i)
- (4 -2i)(6 +5i)
with sign and order variations.
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Comment on trial and error
Actually, we did an exhaustive search of the 441 products of single-digit numbers [-9, 9] to see which pairs of them differed by 34. Then, among those, we looked for product pairs that added to 8. In the end, we found the 8 solutions described above.